About the Author
Stan Gibilisco is an electronics engineer, researcher, and mathematician who has authored a number of titles for the McGraw-HillDemystified series, along with more than 30 other books and dozens of magazine articles. His work has been published in several languages.

2 The Language of Sets 19 The Concept of a Set 19 How Sets Relate 22 Set Intersection 27 Set Union 30 Practice Exercises 33 3 Natural Numbers and Integers 35 How Natural Numbers are Made 35 Special Natural Numbers 38 Natural Number Nontrivia 42 The Integers 45 Practice Exercises 49

vii

viii Contents

4 Addition and Subtraction Moving Up and Down 51

51

Identity, Grouping, and Signs 55 The Commutative Law for Addition 57 The Associative Law for Addition 59 Practice Exercises 63

5 Multiplication and Division 65 Moving Out and In 65 Identity, Grouping, and Signs 70
The Commutative Law for Multiplication 73 The Associative Law for Multiplication 75 The Distributive Laws 78 Practice Exercises 81

you’ll come up with thirteen days if you get it right. You might use a calendar and point to the days one at a time as you count them out. Numerals are tailor-made for solving number problems because they make abstract things easy to envision. thirty-one. twenty-seven. How about ten thousand? You could stick another T onto the left-hand end of the symbol for a thousand. You can represent fifty by using three toothpicks to make a capital letter F. If you attack this problem as I would. You might use objects such as toothpicks to create numerals in a system that expands on this idea. you’ll count out loud starting with tomorrow. four.
Toothpicks on the table Everyone has used “hash marks” to tally up small numbers. two. I would use my fingers to count along or make “hash marks” on a piece of paper (Fig. five. You’re making sure that you get the right answer by using numerals to represent the numbers. You can represent one item by a single mark and five items by four marks with a long slash. That’s what mathematicians did when they invented numeration systems in centuries long past. as shown in Fig. twenty-six. Don’t be embarrassed if you find yourself figuring out simple problems like this using your fingers or other convenient objects. or you could run two letters H together to indicate that it’s a hundred hundred! Use your imagination. However you do it. twenty-eight. You can run a T and an H together to make a symbol that represents a thousand. three. You can represent a hundred by making a capital letter H with three toothpicks. twenty-nine. But be careful! This sort of problem is easy to mess up. This lets you express rather large numbers such as seventy-four or two hundred fifty-three without having to buy several boxes of toothpicks and spend a lot of time laying them down. thirty. running an F and H together to create a symbol that represents five hundred. In this system. any particular arrangement of sticks is a numeral. You can represent ten by making a capital letter T with two toothpicks. one. six!” While jabbering away. 1-2.
A calculator won’t work very well to solve this problem. Try it and see! You can’t get the right answer by any straightforward arithmetic operation on twenty-four and six. You can keep going this way.4 Counting Methods (24) July 25 26 27 28 29 30 31
1 August
2
3
4
5
6
Figure 1-1 How many days pass from the afternoon of July 24
until the afternoon of August 6? You can make marks on a piece of paper and then count them to figure out the answer.
. 1-1). July twenty-fifth (under your breath): “twenty-five.

meaning that it uses the smallest possible number of toothpicks. It’s awkward. most numbers can be represented by more than one numeral. five hundred and so on. It’s impractical for expressing gigantic numbers. and then from ten to a hundred.
Are you confused?
If the toothpick numeral system puzzles you. You can represent seven hundred seventy-seven in more ways than just the two shown here. then to ten thousand.
Solution
Figure 1-3 shows two ways you can represent this number. The arrangement on top is the most “elegant” possible numeral. and so on. But using blocks of five. In this system. conserves toothpicks. But there is always a “best numeral” that uses the smallest possible number of toothpicks.
Here’s a challenge!
Using toothpick numerals represent the number seven hundred seventy-seven in two different ways.Fingers and Sticks
1 5 10 10
5
50
74
100
253
Figure 1-2 Toothpick-numeral equivalents
of some numbers. You can make numerals that are far more “inelegant” than the bottom arrangement. fifty. then to a thousand. don’t feel bad. It’s easier to go straight from blocks of one to blocks of ten. in addition to the traditional multiples of ten. The worst possible approach is to lay down seven hundred seventy-seven toothpicks side-by-side. People aren’t used to counting in blocks of five or fifty or five hundred. you build the F and the H together so they’re a single connected pattern of sticks. Make sure one of your arrangements is the most “elegant” possible way to represent seven hundred seventyseven. In order to represent five hundred.
.

let alone hash marks. For example. Arranging the symbols The people who designed the Roman system did not like to put down more than three identical symbols in a row. There’s a more serious issue. as you lay down the sticks in the toothpick system. this looks like a refinement of the toothpick numeration scheme. Instead of MDXXXX to represent one thousand nine hundred. indicating that the smaller quantity should be taken away from the larger. instead of IIII (four ones) to represent four. But there are some subtle differences. a quantity of one is represented by a capital letter I.
Basic symbols In Roman numerals. and a thousand is usually represented by a capital M. “No wonder people got away from Roman numerals. you’d write MCM (a thousand and then another thousand with a hundred taken away). you would write IV (five with one taken away). When you start adding and subtracting. A quantity of ten is denoted as a capital X. Instead of putting four identical symbols one after another. a hundred is a capital C. five hundred is a capital D. They’re confusing!” But that’s not the only trouble with the Roman numeral system or the toothpick numeral system we made up earlier. Neither of these schemes give you any way to express the quantity zero. Instead of XXXX (four tens) to represent forty. What about zero? By now you must be thinking. Why make a big fuss over a symbol that represents nothing? Sometimes the best way to see why something is important is to try to get along without it. The top method is preferred because it is more “elegant. There are exceptions. you’d write XL (fifty with ten taken away). A quantity of five is represented by a capital V.6 Counting Methods
Figure 1-3 Two different ways of expressing seven hundred
seventy-seven in toothpick numerals.) So far. (Sometimes K is used instead.”
Roman Numerals
The toothpick numeration system just described bears a resemblance to another system that was actually used in much of the world until a few centuries ago: the Roman numeration system. fifty is a capital L. You don’t always write the symbols in straightforward order from left to right. the writer would jump up to the next higher symbol and then put the next lower one to its left. more often called Roman numerals. This might not seem important at first thought. and especially when you start multiplying
. intended to save symbols.

which means ten with nine more added on. and III means three. VII means seven. In a third column. You should also begin to understand why mathematicians abandoned this system centuries ago.
1=I 2 = II 3 = III 4 = IV 5=V 6 = VI 7 = VII 8 = VIII 9 = IX 10 = X 20 = XX 30 = XXX 40 = XL 50 = L 60 = LX 70 = LXX 80 = LXXX 90 = XC 100 = C 200 = CC 300 = CCC 400 = CD 500 = D 600 = DC 700 = DCC 800 = DCCC 900 = CM 910 = CMX 920 = CMXX 930 = CMXXX 940 = CMXL 950 = CML 960 = CMLX 970 = CMLXX 980 = CMLXXX 990 = CMXC 991 = CMXCI 992 = CMXCII 993 = CMXCIII 994 = CMXCIV 995 = CMXCV 996 = CMXCVI 997 = CMXCVII 998 = CMXCVIII 999 = CMXCIX
. we write XIV. To represent nine. Now for fourteen. The first three are easy: the symbol I means one. the presence or absence of a single 0 on a piece of paper can represent the difference between the price of a car and the price of a house. Continuing. Then for four. and XIII means thirteen. XII means twelve.Roman Numerals
7
and dividing. and XXI to represent twenty-one. we write IX. In a computer. VI means six. meaning that one is taken away from five. “Normal” numerals are shown along with their Roman equivalents for clarification. In accounting. This will give you a “feel” for how the symbols are arranged to represent adding-on or taking-away of quantities. and the fifth column is farthest to the right. you should be able to see how the system works for fairly large numbers. Some examples of Roman numerals.
Solution
Refer to Table 1-1. II means two. X means ten.
Are you confused?
Let’s write down all the counting numbers from one to twenty-one as Roman numerals. Then going on. From this progression. XI means eleven. In a fifth column. read downward. it’s almost impossible to get along without zero. we write XIX. and VIII means eight. In a second column. put down the equivalents of one to nine in steps of one.
Here’s a challenge!
Write down some Roman numerals in a table as follows. For increasing values in each column. which means ten with four more added on. put down the equivalents of ten to ninety in steps of ten. Then XV means fifteen. we have XX that stands for twenty. V means five. XVI means sixteen. The first column is farthest to the left. and XVIII means eighteen. put down the equivalents of nine hundred ninety-one to nine hundred ninety-nine in steps of one.
Table 1-1. the numeral 0 is one of only two possible digits (the other being 1) for building large numerals. put down the equivalents of one hundred to nine hundred in steps of one hundred. In the first column. For nineteen. put down the equivalents of nine hundred ten to nine hundred ninety in steps of ten. Proceeding. we write IV. meaning that one is taken away from ten. In a fourth column. XVII means seventeen.

depending on where it was written in relation to other digits in the same numeral. That’s where the digit 0 became useful. 6. 3. and the “Arabic” part from the Middle East. You will often hear this scheme called simply Arabic numerals. 5.8 Counting Methods
Hindu-Arabic Numerals
The numeration system we use today was invented in the seventh century by mathematicians in Southern Asia. The idea was that every digit in a numeral should have ten times the value of the digit (if any) to its right. Zero as a “placeholder” Figure 1-4 shows an example of a numeral that represents a large number. and 9. 2. When building up the numeric representation for a large number. but a definite need for one on either side of it.
The idea of “place” In an Arabic numeral. 7. Note that the digit 0. large numbers are
represented by building up numerals digit-by-digit from right to left. The original Hindu inventors of the system came up with an interesting way of expressing numbers larger than nine.
Multiple of one
Multiple of ten
. most of the civilized world adopted the Hindu-Arabic numeration system. giving each succeeding digit ten times the value of the digit to its right. 1. every digit represents a quantity ranging from nothing to nine. During the next two or three hundred years. These digits are the familiar 0. also called a cipher. is just as important as any other digit. Good ideas have a way of catching on. They gave each digit more or less “weight” or value. The quantity shown is
Multiple of a hundred thousand Multiple of a thousand million
Multiple of a hundred million
Multiple of ten thousand
Multiple of a thousand
Multiple of ten million
Multiple of a million
Multiple of a hundred
0
0
0
0
7
0
8
0
6
5
5 “ones” 6 “tens” No “hundreds” 8 “thousands” No “ten thousands” 7 “hundred thousands” No higher-valued digits
Figure 1-4 In the Hindu-Arabic numeration system. there would occasionally be no need for a digit in a particular place. 4. invaders from the Middle East picked it up. The “Hindu” part of the name comes from India. even with invading armies! Eventually. 8.

• The numeral 1 followed by thirty-three ciphers represents a decillion. and it is unusual to see any of them written down. Once you get to a certain nonzero digit as you work your way from right to left. we deal with numbers that make a thousand seem tiny by comparison.) It’s customary to place a comma or space after every third digit as you proceed from right to left in a multi-digit numeral like this. all of them gray (just to remind you that each of them is there in theory. Usage varies depending on which text you happen to read. • The numeral 1 followed by fifteen ciphers represents a quadrillion.
. and so on. • The numeral 1 followed by twenty-seven ciphers represents an octillion. • The numeral 1 followed by thirty ciphers represents a nonillion.
Counting vs. (Some people would call it seven hundred and eight thousand and sixty-five. 1-4 by dozens of places. But in today’s scientific world. three. four. If you travel to the left of the digit 7 in Fig. 1-4. two. making it clear what the values of digits to its left should be. three. Here are some of the names for numbers that are represented as a 1 followed by multiples of three ciphers:
• The numeral 1 followed by three ciphers represents a thousand. all the digits farther to the left are understood to be ciphers. All those ciphers to the left of the last nonzero digit are insignificant in most situations.Hindu-Arabic Numerals
9
seven hundred eight thousand sixty-five. the counting numbers go as one. A simple change in a numeral can make a big difference in the number it represents. and then change one of those ciphers to the digit 1. They can be defined with the Roman numeration system. For our purposes. The only difference is whether we start at one or zero. But once in a while you might find it helpful to insert one or more of them during a calculation. one.
How many numbers exist? Envision an endless string of ciphers continuing off to the left in Fig. Names for some huge numbers People who used Roman numerals hardly ever had to work with numbers much larger than a thousand. • The numeral 1 followed by nine ciphers represents a billion in the United States or a thousand million in England. • The numeral 1 followed by eighteen ciphers represents a quintillion • The numeral 1 followed by twenty-one ciphers represents a sextillion. • The numeral 1 followed by six ciphers represents a million. Every digit 0 “inside” a numeral serves as a placeholder. • The numeral 1 followed by twelve ciphers represents a trillion in the United States or a billion in England. • The numeral 1 followed by twenty-four ciphers represents a septillion. This is an example of the power of the Arabic numeration system. passing through 0 after 0. whole numbers Let’s make sure we understand the difference between a counting number and a whole number. five. waiting to be changed to some other digit if you need to express a huge number). two. and can also be defined with the toothpick system we “invented” here. the value of the represented number increases fantastically. four. and so on. We’ll define the whole numbers as zero. five.

. The original inventors of the Arabic system put down a dot or a tiny circle instead of the full-size digit 0. and you get the representation for a larger whole number. 78. “infinity” can be represented by a lemniscate (∞). But there is no limit to the number of whole numbers you can denote that way. Don’t forget to include the commas where they belong! For example: • • • • 700 represents a quantity that’s a hundred times as large as 7.400 represents a quantity that’s a hundred times as large as 14.” Why bother with commas or spaces. The group. 1. or the capital Hebrew letter aleph (ℵ) with a numeric subscript that defines its “density. Even if a string of digits is hundreds of miles long. 1.900 represents a quantity that’s a hundred times as large as 789. as people did when the Roman system ruled. Mathematicians have found more than one type of “infinity”! Depending on the context.10 Counting Methods
Another interesting property of the Arabic system is the fact that there is no limit to how large a numeral you can represent. That means there is no largest whole number.
What about “infinity”? That elusive thing we call “infinity” is entirely different from any whole number. But the cipher and the comma (or space) make errors a lot less likely.400. Mathematicians use the term finite to describe anything that ends somewhere. or set.000. You don’t have to keep inventing new symbols when numbers get arbitrarily large. the Arabic system lets you do it in a finite number of digits. even if it circles the earth. even if it goes from the earth to the moon—all you have to do is put a nonzero digit on the left or any digit on the right. it represents “nothing.
Here’s a challenge!
Imagine a whole number represented by a certain string of digits in the Arabic system. no matter what the digits happen to be?
Solution
You can make any counting numeral stand for a number a hundred times as large by attaching two ciphers to its right-hand end. Try it with a few numerals and see. Every imaginable number can be represented as an Arabic numeral that contains a finite number of digits. How can you change the Arabic numeral to make the number a hundred times as large.”
Are you confused?
Do you still wonder why the digit 0 is needed? After all. No matter how large a whole number you want to express. the small Greek letter omega (ω). or any other sort of number people usually imagine. either? The quick answer to these questions is that the digit 0 and the comma (or space) are not actually needed in order to write numerals. of all whole numbers is said to be infinite (not finite).000 represents a quantity that’s a hundred times as large as 14. and every single one of those digits is from the basic set of 0 through 9.

The radix-ten system. the fourth line as C. 8 always mean the quantity eight. just hold on a couple of minutes!) When the expression for a number is a spelled-out word like “eighteen” or “forty-five” or “three hundred twenty-one. at least when you start counting in it.” you must know the radix to be sure of how many apples the basket contains. and 16 always mean the quantity sixteen? Not necessarily! It’s true in baseten. “There are forty-five apples in this basket. 1-5B. but it is not necessarily true in other bases. 1-5A. The base-two and base-sixteen systems.” it is absolutely clear what I mean. not with 0. In the base-sixteen numeration system. not counting commas (or decimal points. keeping the 1 in the tens place. place a digit 1 to the left of the 0 and then go around again. Then you must change the tens digit back to 0 and place a 1 in the hundreds place. the total number of pound signs in the third line would be written as the letter A. in particular.
A subtle distinction Doesn’t 5 always mean the quantity five. also called base-ten or the decimal numeration system. regardless of the radix. You start with I (which stands for the number one). If I write. 10 always mean the quantity ten. change the tens digit to 2. therefore has ten symbols. You can keep going this way until you have completed the tenth revolution in which you have a 9 in the tens place. “There are 45 apples in this basket.” we mean the actual quantity. When you have completed the second revolution. imagine proceeding clockwise around the face of a ten-hour clock as shown in Fig. You can complete one revolution and go through part of the second and the system works well.The Counting Base
11
The Counting Base
The radix or base of a numeration system is the number of single-digit symbols it has. When you have completed the first revolution.
The decimal system As you count upward from zero in the base-ten system. (If you’re confused right now. the fourth line as 14. the total number of pound signs in the second line in the above list would be written as 10. the third line as 12.
. Imagine a five-hour clock such as the one shown in Fig. You can get a good “mental workout” by playing with these! But they’re more than mind games.
• • • • • Here are five pound signs: ##### Here are eight pound signs: ######## Here are ten pound signs: ########## Here are twelve pound signs: ############ Here are sixteen pound signs: ################
In the base-eight numeration system. But if I write. The Roman system The Roman numeration scheme can be considered as a base-five system. But there are systems that use bases other than ten. In this section we’ll look at some of them. are commonly used in computer science. and the last line as 10. and that have more or less than ten symbols to represent the digits. which we’ll get into later). and the last line as 20.

It indicates that a pattern continues for a while. The orderliness of this system falls apart before you even get twice around!
The octal system Now imagine an eight-hour clock as shown in Fig.. 7. place a 1 to the left of the digits shown. VIII. 11. Roman base-five (B). The number nine is not represented as VIV.. proceed clockwise. so you count
. 12. But skip the digits 8 and 9. but X is not on the clock face. But when you get past VIII (which stands for the number eight). . It’s written as IX. They do not exist in this system.
After the first revolution. Use the same upward-counting scheme as you did with the ten-hour clock. VII. 5. you keep the V and then start adding symbols to its right: VI. As you count. 1-5C. a problem occurs.. base-eight or octal (C). This shows how the base-eight or octal numeration system works. (You’ll
. or perhaps even forever. 10. 6. saving you from having to do a lot of symbol scribbling.. although technically it could be. and base-sixteen or hexadecimal (D).12 Counting Methods 0 I 2 3 6 5 A 4 IV III V II
9 8 7
1
B
0 7 6 5 4 C 3 1 2
E D C B A
F 0 1
2 3 4 5
9 8 7 D
6
Figure 1-5 Clock-like representations of digits in base-ten or
decimal (A). The string of three dots is called an ellipsis. When you finish the first revolution and are ready to start the second.

. 9F.. as shown in Fig.. you count . FF. When you finish up the eighth revolution and enter the ninth. 1D. 1C. 100. 21. . 1B. You count . D. you count . 20. 9D. You can see from this drawing how the base-sixteen or hexadecimal numeration system works. 1F. 101.. A0.. 1A. 98.. 13. 76.
The hexadecimal system Let’s invent one more strange clock. 19. Then you get to the end of the sixteenth revolution and move into the seventeenth. 21. 102.. . 100. B. E. When you complete the tenth revolution and move into the eleventh. .. FC. in addition to the digits in the base-ten system:
• • • • • • A stands for ten B stands for eleven C stands for twelve D stands for thirteen E stands for fourteen F stands for fifteen
When you finish the first revolution and move into the second. 16. 103... FD. 22.The Counting Base
13
see this notation often in mathematics. 20. There are six new digits here. 75. 9B. 8. D. place a l to the left of the digits shown. F. It goes on like this with B.. C. you count . A2.. .. 1E. F9.. C.. 9E. 1-5D. E. 9A.. Continuing through the second revolution and into the third.. FE.. and F in the sixteens place. 99. A1. FA. 23. 11. 22. A3. you count . .. 102. 77. 18. . A. 17.) Continuing through the second revolution and into the third.. 10. 101. Use the same upward-counting scheme as you did with the ten-hour and eight-hour clocks. This one has sixteen hours. Get the idea?
. 15. 9C.. 12. FB. 9.. like this: F8.

14 Counting Methods
The binary system When engineers began to design electronic calculators and computers in the twentieth century. they wanted a way to count up to large numbers using only two digits.
Multiple of two hundred fifty-six
Multiple of five hundred twelve
Multiple of one hundred twenty-eight
Multiple of sixty-four
Multiple of thirty-two
Multiple of sixteen
Multiple of eight
Multiple of four
0
0
0
0
1
0
1
0
1
1 “one” 1 “two” No “fours” 1 “eight” No “sixteens” 1 “thirty-two” No higher-valued digits
Figure 1-6 In the binary system. high-speed electronic switches. the machine converts it into a binary numeral.
Multiple of one 1
Multiple of two
. one to represent the “off ” condition of an electrical switch and the other to represent the “on” condition.” “low/high.” “no/yes. giving each succeeding digit twice the value of the digit to its right. Instead of going up by multiples of ten. The result is a base-two or binary numeration system. Note the absence of commas in this system. Every binary numeral has a unique equivalent in the decimal system. When you use a computer or calculator and punch in a series of decimal digits. eight. but binary numerals can be easily represented by the states of simple. or sixteen. These two states can also be represented as “false/true. Numerals in the binary system are longer than numerals in the other systems. Figure 1-6 shows how numerals in the binary system are put together. and vice versa. performs whatever calculations or operations you demand. large numbers are represented by
building up numerals digit-by-digit from right to left. you double the value of each digit as you move one place to the left.” or as the numerals 0 and 1.” “negative/positive.

523
One more thing .The Counting Base
15
converts the result back to a decimal numeral. • In the ones place you have 3. Note that D represents thirteen. so you start out with that • In the sixteens place you have 0.
Are you confused?
Table 1-2 compares numerical values in the base-ten. billions. which is three thousand three hundred twenty-eight. so you must add thirteen times two hundred fifty-six.
Solution
To solve this. Fortunately. We won’t have to worry about ambiguity that could result from an alternative radix such as eight or sixteen. which is zero.” That’s one of the reasons why numerals were invented!
. all of the electronic switching actions and manipulations take place out of your sight. you can figure it out as follows. you are finished at this point. you should be able to figure out (with a little bit of thought and scribbling) how to convert larger decimal numerals to any of the other forms. All of the conversions and calculations. so you must add two times four thousand ninety-six. Don’t use a computer or go on the Internet to find a Web site that will do it for you. base-eight. Are you getting tired of reading numbers as words? In the rest of this book. to what you have so far • In the two hundred fifty-sixes place you have D which means thirteen. just itself ). to what you have so far Because there are no digits to the left of the 2. which is eight thousand one hundred ninety-two. expressed as a sum in decimal numerals.192 = 11. and base-sixteen systems from zero to sixty-four. and then displays that numeral for you. at incredible speed.. The final result.328 + 8. to what you have so far • In the four thousand ninety-sixes place you have 2. The digit farthest to the right represents a multiple of one (that is. Grind it out the long way.. so you must add zero times sixteen. Numerals will also come in handy when numbers get large or “messy. base-two. The next digit to the left represents a multiple of sixteen. and trillions!
Here’s a challenge!
Convert the hexadecimal numeral 2D03 to decimal form. From this table. Then comes a multiple of four thousand ninety-six (sixteen times two hundred fifty-six). So we’ll start using numerals to represent specific quantities most of the time. is 3 + 0 + 3. there are plenty of computer programs and Web sites that will do such conversions for you up to millions. After that comes a multiple of two hundred fifty-six (or sixteen times sixteen). Thinking in the decimal system. you need to know the place values. we’ll be dealing in the decimal system exclusively.

Convert the following Roman numerals to decimal numerals. Write down the number three hundred two trillion. What numeral in the octal (base-eight) system represents the same quantity that the binary numeral shown in Fig. 1-6 represents?
. Consider a billion as a thousand million. one hundred ten as a decimal numeral. How can you make the number represented by the numeral in the answer to Problem 4 ten times as large? A hundred times as large? A thousand times as large? 7. (a) MMXX (b) MMXIX (c) MMIX (d) MMVI 4. six thousand.18 Counting Methods
3. 1-6 represents? 9. How many ciphers could you add to the left of the digit 3 in the decimal numeral in the situation of Problem 4 without changing the value of the number it represents? 6. Include commas where appropriate. seventy billion. and a trillion as a million million. 1-6 represents? 10. What numeral in the hexadecimal (base-sixteen) decimal system represents the same quantity that the binary numeral shown in Fig. 5. What numeral in the decimal (base-ten) system represents the same quantity that the binary numeral shown in Fig. one hundred fortynine million. How can you write out the final answer to Problem 6 as a number in words rather than as a numeral in digits? 8.

Listing the elements When the elements of a set are listed, the list is enclosed in “curly brackets,” usually called braces. The order of the list does not matter. Repetition doesn’t matter either. The following sets are all the same:
{1, 2, 3} {3, 2, 1} {1, 3, 3, 2, 1} {1, 2, 3, 1, 2, 3, 1, 2, 3, ...} The ellipsis (string of three dots) means that the list goes on forever in the pattern shown. In this case, it’s around and around in an endless cycle. Now look at this example of a set with five elements: S = {2, 4, 6, 8, 10} Are the elements of this set S meant to be numbers or numerals? That depends on the context. Usually, when you see a set with numerals in it like this, the author means to define the set containing the numbers that those numerals represent. Here’s another example of a set with five elements: P = {Mercury, Venus, Earth, Mars, Jupiter} You’re entitled to assume that the elements of this set are the first five planets in our solar system, not the words representing them.

The empty set A set can exist even if there are no elements in it. This is called the empty set or the null set. It can be symbolized by writing two braces facing each other with a space between, like this:
{} Another way to write it is to draw a circle and run a forward slash through it, like this: ∅ Let’s use the circle-slash symbol in the rest of this chapter, and anywhere else in this book the null set happens to come up. You might ask, “How can a set have no elements? That would be like a club with no members!” Well, so be it, then! If all the members of the Pingoville Ping-Pong Club quit today and no new members join, the club still exists if it has a charter and by laws. The set of members of the Pingoville Ping-Pong Club might be empty, but it’s a legitimate set as long as someone says the club exists.

Finite or infinite? Sets can be categorized as either finite or infinite. When a set is finite, you can name all of its elements if you have enough time. This includes the null set. You can say “This set has no

The Concept of a Set

21

elements,” and you’ve named all the elements of the null set. When a set is infinite, you can’t name all of its elements, no matter how much time you have. Even if a set is infinite, you might be able to write an “implied list” that reveals exactly what all of its elements are. Consider this: W = {0, 1, 2, 3, 4, 5, ...} This is the set of whole numbers as it is usually defined in mathematics. You know whether or not something is an element of set W, even if it is not shown above, and even if you could not reach it if you started to scribble down the list right now and kept at it for days. You can tell right away which of the following numbers are elements of W, and which are not: 12 1/2 23 100/3 78,883,505 356.75 90,120,801,000,000,000 −65,457,333 The first, third, fifth, and seventh numbers are elements of W, but the second, fourth, sixth, and eighth numbers are not. Some infinite sets cannot be totally defined by means of any list, even an “implied list”! You’ll learn about this type of set in Chap. 9.

Sets within sets A set can be an element of another set. Remember again, anything can be a member of a set! You can have sets that get confusing when listed. Here are some examples, in increasing order of strangeness:
{1, 2, 3, 4, 5} {1, 2, {3, 4, 5}} {1, {2, {3 ,4, 5}}} {1, {2, {3, {4, 5}}}} {1, {2, {3, {4, {5}}}}} An “inner” or “member” set can sometimes have more elements than the set to which it belongs. Here is an example: {1, 2, {3, 4, 5, 6, 7, 8}} Here, the main set has three elements, one of which is a set with six elements.

22 The Language of Sets

Are you confused?
Do you still wonder what makes a bunch of things a set? If you have a basket full of apples and you call it a set, is it still a set when you dump the apples onto the ground? Were those same apples elements of a set before they were picked? Questions like this can drive you crazy if you let them. A collection of things is a set if you decide to call it a set. It’s that simple. As you go along in this course, you’ll eventually see how sets are used in algebra. Here’s an easy example. What number, when multiplied by itself, gives you 4? The obvious answer is 2. But −2 will also work, because “minus times minus equals plus.” In ordinary mathematics, a number can’t have more than one value. But two or more numbers can be elements of a set. A mathematician would say that the solution set to this problem is {−2, 2}.

Here’s a riddle!
You might wonder if a set can be an element of itself. At first, it is tempting to say “No, that’s impossible. It would be like saying the Pingoville Ping-Pong Club is one of its own members. The elements are the Ping-Pong players, not the club.” But wait! What about the set of all abstract ideas? That’s an abstract idea. So a set can be a member of itself. This is a strange scenario because it doesn’t fit into the “real world.” In a way, it’s just a riddle. Nevertheless, riddles of this sort sometimes open the door to important mathematical discoveries.

Here’s a challenge!
Define the set of all the positive and negative whole numbers in the form of an “implied list” of numerals. Make up the list so that, if someone picks a positive or negative number, no matter how big or small it might be, you can easily tell whether or not it is in the set by looking at the list.

Solution
You can do this in at least two ways. You can start with zero and then list the numerals for the positive and negative whole numbers alternately: {0, 1, −1, 2, −2, 3, −3, 4, −4, ...} You can also make the list open at both ends, implying unlimited “travel” to the left as well as to the right: {..., −4, −3, −2, −1, 0, 1, 2, 3, 4, ...}

How Sets Relate
Now let’s see how sets can be broken down, compared, and combined. Pictures can do the work of thousands of words here.

Venn diagrams One of the most useful illustrations for describing relationships among sets is a Venn diagram, in which sets are shown as groups of points or as geometric figures. Figure 2-1 is an example. The

How Sets Relate
Universe All the positive whole numbers

23

4

2 6 1/8

1/10 1/12 All the people in Illinois All the negative whole numbers All the women in Chicago

Figure 2-1 A Venn diagram showing the set of all sets (the
universe) along with a few specific sets within it.

large, heavy rectangle represents the set of all things that can exist, whether real or imaginary (and that includes all possible sets). This “emperor of sets” is called the universal set or the universe. In Fig. 2-1, three of the sets shown inside the universe are finite and two are infinite. Note how the objects overlap or are contained within one another or are entirely separate. This is important, because it describes the various ways sets can relate to each other. You can see how this works by examining the diagram carefully. All the women in Chicago are people in Illinois, but there are plenty of people in Illinois who aren’t women in Chicago. The numbers 2, 4, and 6 are positive whole numbers, but there are lots of positive whole numbers different from 2, 4, or 6. The sets of positive and negative whole numbers are entirely separate, even though both sets are infinite. None of the positive or negative whole numbers is a person in Illinois, and no person in Illinois is number (except according to the government, maybe).

Subsets When all the elements of a set are also contained in a second set, the first set is called a subset of the second. If you have two sets A and B, and every element of A is also an element of B, then A is a subset of B. That fact can be written
A⊆B

24 The Language of Sets

Figure 2-1 shows that the set of all the women in Chicago is a subset of the set of all the people in Illinois. That is expressed by a hatched square inside a shaded oval. Figure 2-1 also shows that the set {2, 4, 6} is a subset of the set of positive whole numbers. That is expressed by placing the numerals 2, 4, and 6 inside the rectangle representing the positive whole numbers. All five of the figures inside the large, heavy rectangle of Fig. 2-1 represent subsets of the universe. Any set you can imagine, no matter how large, small, or strange it might be, and no matter if it is finite or infinite, is a subset of the universe. Technically, a set is always a subset of itself. Often, a subset represents only part, not all, of the main set. Then the smaller set is called a proper subset of the larger one. In the situation shown by Fig. 2-1, the set of all the women in Chicago is a proper subset of the set of all the people in Illinois. The set {2, 4, 6} is a proper subset of the set of positive whole numbers. All five of the sets inside the main rectangle are proper subsets of the universe. When a certain set C is a proper subset of another set D, we write C ⊂D

Congruent sets Once in a while, you’ll come across two sets that are expressed in different ways, but they turn out to be exactly the same when you look at them closely. Consider these two sets:
E = {1, 2, 3, 4, 5, ...} F = {7/7, 14/7, 21/7, 28/7, 35/7, ...} At first glance, these two sets look completely different. But if you think of their elements as numbers (not as symbols representing numbers) the way a mathematician would regard them, you can see that they’re really the same set. You know this because 7/7 = 1 14/7 = 2 21/7 = 3 28/7 = 4 35/7 = 5 ↓ and so on, forever Every element in set E has exactly one “mate” in set F, and every element in set F has exactly one “mate” in set E. In a situation like this, the elements of the two sets exist in a one-to-one correspondence. When two sets have elements that are identical, and all the elements in one set can be paired off one-to-one with all the elements in the other, they are said to be congruent sets. Sometimes they’re called equal sets or coincident sets. In the above situation, we can write E=F

How Sets Relate

25

Once in a while, you’ll see a three-barred equals sign to indicate that two sets are congruent. In this case, we would write E≡F

Disjoint sets When two sets are completely different, having no elements in common whatsoever, then they are called disjoint sets. Here is an example of two disjoint sets of numbers:
G = {1, 2, 3, 4} H = {5, 6, 7, 8} Both of these sets are finite. But infinite sets can also be disjoint. Take the set of all the even whole numbers and all the odd whole numbers: Weven = {0, 2, 4, 6, 8, ...} Wodd = {1, 3, 5, 7, 9, ...} No matter how far out along the list for Weven you go, you’ll never find any element that is also in Wodd. No matter how far out along the list for Wodd you go, you’ll never find any element that is also in Weven. We won’t try to prove this right now, but you should not have any trouble sensing that it’s a fact. Sometimes the mind’s eye can see forever! Figure 2-2 is a Venn diagram showing two sets, J and K, with no elements in common. You can imagine J as the set of all the points on or inside the circle and K as the set of all the points on or inside the oval. Sets J and K are disjoint. When you have two disjoint sets, neither of them is a subset of the other.

Universe

J

K

Figure 2-2 Two disjoint sets, J and K. They have no
elements in common.

26 The Language of Sets

Overlapping sets When two sets have at least one element in common, they are called overlapping sets. In formal texts you might see them called nondisjoint sets. Congruent sets overlap in the strongest possible sense, because they have all their elements in common. More often, two overlapping sets share some, but not all, of their elements. Here are two sets of numbers that overlap with one element in common:
L = {2, 3, 4, 5, 6} M = {6, 7, 8, 9, 10} Here is a pair of sets that overlap more: P = {21, 23, 25, 27, 29, 31, 33} Q = {25, 27, 29, 31, 33, 35, 37} Technically, these sets overlap too: R = {11, 12, 13, 14, 15, 16, 17, 18, 19} S = {12, 13, 14} Here, you can see that S is a subset of R. In fact, S is a proper subset of R. Now, let’s look at a pair of infinite sets that overlap with four elements in common: W3− = {..., −5, −4, −3, −2, −1, 0, 1, 2, 3} W0+ = {0, 1, 2, 3, 4, 5, ...} The notation W3− (read “W sub three-minus”) means the set of all positive or negative whole numbers starting at 3 and decreasing, one by one, without end. The notation W0+ (read “W sub zero-plus”) means the set of whole numbers starting with 0 and increasing, one by one, without end. That’s the set of whole numbers as it is usually defined. Figure 2-3 is a Venn diagram that shows two sets, T and U, with some elements in common, so they overlap. You can imagine T as the set of all the points on or inside the circle, and
Universe

T

U

Figure 2-3 Two overlapping sets, T and U. They
have some elements in common.

Set Intersection

27

U as the set of all the points on or inside the oval. When you have two overlapping sets, one of them can be a subset of the other, but this does not have to be the case. It is clearly not true of the two sets T and U in Fig. 2-3. Neither of these sets is a subset of the other, because both have some elements all their own.

Set Intersection
The intersection of two sets is made up of all elements that belong to both of the sets. When you have two sets, say V and W, their intersection is also a set, and it is written V ∩ W. The upside-down U-like symbol is read “intersect,” so you would say “V intersect W.”

Intersection of two congruent sets When two nonempty sets are congruent, their intersection is the set of all elements in either set. You can write it like this, for any nonempty sets X and Y,
If X = Y then X∩Y=X and X∩Y=Y But really, you’re dealing with only one set here, not two! So you could just as well write X∩X=X This also holds true for the null set: ∅∩∅=∅

Intersection with the null set The intersection of the null set with any nonempty set gives you the null set. This fact is not so trivial. You might have to think awhile to fully understand it. For any nonempty set V, you can write
V∩∅=∅ Remember, any element in the intersection of two sets has to belong to both of those sets. But nothing can belong to a set that contains no elements! Therefore, nothing can belong to the intersection of the null set with any other set.

28 The Language of Sets

Intersection of two nonempty disjoint sets When two nonempty sets are disjoint, they have no elements in common, so it’s impossible for anything to belong to them both. The intersection of two disjoint sets is always the null set. It doesn’t matter how big or small the sets are. Remember the sets of even and odd whole numbers, Weven and Wodd? They’re both infinite, but
Weven ∩ Wodd = ∅

Intersection of two overlapping sets When two sets overlap, their intersection contains at least one element. There is no limit to how many elements the intersection of two sets can have. The only requirement is that every element in the intersection set must belong to both of the original sets. Let’s look at the examples of overlapping sets you saw a little while ago, and figure out the intersection sets. First, examine these
L = {2, 3, 4, 5, 6} M = {6, 7, 8, 9, 10} Here, the intersection set contains one element: L ∩ M = {6} That means the set containing the number 6, not just the number 6 itself. Now look at these: P = {21, 23, 25, 27, 29, 31, 33} Q = {25, 27, 29, 31, 33, 35, 37} The intersection set in this case contains five elements: P ∩ Q = {25, 27, 29, 31, 33} Now check these sets out: R = {11, 12, 13, 14, 15, 16, 17, 18, 19} S = {12, 13, 14} In this situation, S ⊂ R, so the intersection set is the same as S. We can write that down as follows: R∩S=S = {12, 13, 14} How about the set W3− of all positive, negative, or zero whole numbers less than or equal to 3, and the set W0+ of all the nonnegative whole numbers? W3− = {..., −5, −4, −3, −2, −1, 0, 1, 2, 3} W0+ = {0, 1, 2, 3, 4, 5, ...}

Set Intersection Universe V V U W

29

W

Figure 2-4 Two overlapping sets, V and W. Their
intersection is shown by the double-hatched region.

Here, the intersection set has four elements: W3− ∩ W0+ = {0, 1, 2, 3} Figure 2-4 is a Venn diagram that shows two overlapping sets. Think of V as the rectangle and everything inside it. Imagine W as the oval and everything inside it. The two regions are hatched diagonally, but in different directions. The intersection V ∩ W shows up as a double-hatched region.

Are you confused?
Go back and look again at Fig. 2-1. You can see that the set of all women in Chicago (call it Cw) is a proper subset of the set of all people in the state of Illinois (call it Ip). You would write down this fact as follows: Cw ⊂ Ip The diagram also makes it clear that the intersection of set Cw with set Ip is just the set Cw. In order to be in both sets, a person must be a woman in Chicago, that is, an element of Cw. Here’s how you would write that Cw ∩ Ip = Cw You can always draw a Venn diagram if it will help you understand how sets are related.

Here’s a challenge!
Find two sets of whole numbers that overlap, with neither set being a subset of the other, and whose intersection set contains infinitely many elements.

30 The Language of Sets

Solution
There are countless examples of set pairs like this. Let’s look at the set of all positive whole numbers divisible by 4 without a remainder. (When there is no remainder, a quotient comes out as a whole number.) Name this set W4d. Similarly, name the set of all positive whole numbers divisible by 6 without a remainder W6d. Then W4d = {0, 4, 8, 12, 16, 20, 24, 28, 32, 36, ...} W6d = {0, 6, 12, 18, 24, 30, 36, 42, 48, ...} Both of these sets have infinitely many elements. They overlap, because they share certain elements. But neither is a subset of the other, because they both have some elements all their own. Their intersection is the set of elements divisible by both 4 and 6. Let’s call it W4d6d. If you’re willing to write out both of the above lists up to all values less than or equal to 100, you will see that W4d ∩ W6d = W4d6d = {0, 12, 24, 36, 48, 60, 72, 84, 96, ...} This is an infinite set, and it happens to be the set of all positive whole numbers divisible by 12 without a remainder (call it W12d). We can write W4d ∩ W6d = W12d

Set Union
The union of two sets contains all of the elements that belong to one set or the other, or both. When you have two sets, say X and Y, their union is also a set, written X ∪ Y. The U-like symbol is read “union,” so you would say “X union Y.”

Union of two congruent sets When two nonempty sets are congruent, their union is the set of all elements in either set. For any nonempty sets X and Y,
If X = Y then X∪Y=X and X∪Y=Y But you’re really dealing with only one set here, so you could just as well write X∪X=X

Set Union

31

And for the null set ∅∪∅=∅ When two sets are congruent, their union is the same as their intersection. This might seem trivial right now, but there are situations where it’s not clear that two sets are congruent. In cases like that, you can compare the union with the intersection as a sort of congruence test. If the union and intersection turn out identical, then you know the two sets in question are congruent.

Union with the null set The union of the null set with any nonempty set gives you that nonempty set. For any nonempty set X, you can write
X∪∅=X Remember, any element in the union of two sets only has to belong to one of them.

Union of two disjoint sets When two nonempty sets are disjoint, they have no elements in common, but their union always contains some elements. Consider again the sets of even and odd whole numbers, Weven and Wodd. Their union is the set of all the whole numbers. So
Weven ∪ Wodd = {0, 1, 2, 3, 4, 5, ...}

Union of two overlapping sets Again, let’s look at the same examples of overlapping sets we checked out when we worked with intersection. First
L = {2, 3, 4, 5, 6} M = {6, 7, 8, 9, 10} The union set here contains nine elements: L ∪ M = {2, 3, 4, 5, 6, 7, 8, 9, 10} The number 6 appears in both sets, but we count it only once in the union. (An element can only “belong to a set once.”) Now look at these: P = {21, 23, 25, 27, 29, 31, 33} Q = {25, 27, 29, 31, 33, 35, 37} The union set in this case is P ∪ Q = {21, 23, 25, ..., 33, 35, 37}

32 The Language of Sets

That’s all the odd whole numbers between, and including, 21 and 37. We count the duplicate elements 25 through 33 only once. Now look at these: R = {11, 12, 13, 14, 15, 16, 17, 18, 19} S = {12, 13, 14} In this situation, S ⊂ R, so the union set is the same as R. We can write that down this way: R∪S=R = {11, 12, 13, 14, 15, 16, 17, 18, 19} We count the elements 12, 13, and 14 only once. Now these: W3− = {..., −5, −4, −3, −2, −1, 0, 1, 2, 3} W0+ = {0, 1, 2, 3, 4, 5, ...} Here, the union set consists of all the positive and negative whole numbers, along with zero. Let’s write that set as W0± (read “W sub zero plus-or-minus”). Then W3− ∪ W0+ = W0± = {..., −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, ...} The elements 0, 1, 2, and 3 are counted only once. This set W0± is usually called the set of integers. We’ll work more with integers in the chapters to come. Figure 2-5 is a Venn diagram that shows two overlapping sets. Think of X as the rectangle and everything inside it. Imagine Y as the oval and everything inside it. The union of the sets, X ∪ Y, is shown by the entire shaded region inside the outer solid line. Part of that line is the
Universe XUY

X

Y

Figure 2-5 Two overlapping sets, X and Y. Their union is
shown by the entire shaded region.

Practice Exercises

33

outside of the rectangle and part of it is the outside of the oval. Any element inside the region bounded by the dashed line is counted only once.

Are you confused?
Once more, go back and look at Fig. 2-1, again noting that the set of all the women in Chicago is a proper subset of the set of all the people in Illinois, that is, Cw ⊂ Ip. The diagram also makes it plain that the union of Cw with Ip is just Ip. To be in one set or the other (or both), a person only has to be a resident of Illinois, that is, an element of Ip. It’s not necessary to be a woman, and it’s not necessary to be in Chicago. Here’s how you would write that: Cw ∪ Ip = Ip

Here’s a challenge!
Can you find two sets of whole numbers, with one of them infinite, but such that their union contains only a finite number of elements?

Solution
Don’t think about this for too long. You’ll never find two such sets! An element in the union of two sets only has to belong to one of the sets. If a set has infinitely many elements, then the union of that set with any other set—even the null set—must have infinitely many elements as well.

Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems. Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it! 1. Is there any set that is a subset of every other set? If so, what is it? If such a set can’t exist, why not? 2. Continuing with the theme of Problem 1, is there a way to take nothing and build up an unlimited number of different sets from it? If so, show an example. If not, explain why not. 3. What set does the small, dark-shaded triangle marked P represent in Fig. 2-6? What set does the dark-shaded, irregular, four-sided figure marked Q represent? 4. If you consider all the possible intersections of two sets in Fig. 2-6, which of those intersection sets are empty? 5. Is the universal set a subset of itself? Is it a proper subset of itself? 6. Give an example of two sets, both with infinitely many elements, but such that one is a proper subset of the other.

Finally.} G = {1.34 The Language of Sets Universe
B A
P
E
Q
D C
Figure 2-6 Illustration for Probs. 1/6... the denominator of the fraction increases by 1 as you go down the list. List all the subsets of {1.. 2. {2. 8. 1/32. List all the subsets of {1. Here’s a hint: Whenever you want to find all the subsets of a small set like this. 1/4. Be careful. All the numerators in both sets are equal to 1.} In set A. In set G. 3 and 4. 1/4. 1/16. 1/2. {3}}}.
. List all the subsets of {1. 1/5. the denominator doubles as you go down the list. 3}. The hint given with Problem 8 is important here.
7. 1/2. first list its individual elements. 10. . Be extra careful! The hint given with Problem 8 is even more important here. 9. {2. . 1/8. Then make up every possible set that contains at least one of those elements. What is the intersection of these two sets? What is their union? A = {1. which is a subset of any other set. 3}}. be sure to include the empty set.. 1/3.

we can take all the natural numbers up to and including n.. n} If we have any particular natural number p. That allows us to build the set of natural numbers. 1.
Figure 3-2
. 4. which is the same as the set containing the number 0. every element p in the set of natural numbers “rests on” all the natural numbers “below” itself. those elements could be anything.. we can define it on the basis of all the natural numbers less than itself. It has one element. 2.36 Natural Numbers and Integers Start here
and go on forever
Figure 3-1 The number 0 can be defined as the null set. Call it n. atoms. Once this process is defined. look at Fig. n + 1.. 1.. 1. stars. 3.. This drawing shows the first six natural numbers (0. If we want to create the next higher natural number. such as apples. But it’s convenient to use all the natural numbers less than p as those elements. like stacking coins. This is a nifty scheme! A natural number p is a set containing p elements. 2.
0
1 { }
Whole numbers yet to be defined The number 1 is the set containing the null set. 3-3. Just as each coin in the stack rests on all the coins below itself. A mathematician would write it like this:
n + 1 = {0. In theory. We
can show how it starts to generate natural numbers by placing it at the beginning of an endless string of points. one after another. 3.. and 5) as they are built up and assigned to a vertical stack of points that ascends upward as far as we care to go. Suppose we’ve built a certain natural number. it sets in motion a mathematical “chain reaction” that never ends. and then call that set the new number. 3. like this: p = {0. make them into elements of a set. . . 4. 2. or people. one on top of the other.
Building new numbers Let’s make up a rule that we can use to generate new natural numbers. p − 1} If that’s a little too abstract for you.

but if you’re interested in learning more
. 2.
Proceed forever! 5 4 3 2 1 0 {0. 3. What number does it represent. 1. 2.” In formal terms. But the mere fact that we can’t write the whole list doesn’t mean that the set itself does not exist. and it has some strange properties. 1.. 3} {0. ω is called an infinite ordinal or transfinite ordinal. .. each one “on top” of its predecessors. which we called W (for whole numbers) in the last chapter. 2. 2. 2} {0. so in the mathematical world. 1. is symbolized N.. We can write N = {0. 1} which has two elements 3 = {0. . forever Now think of the set of all natural numbers. it exists! Mathematicians define the number represented by the entire set N as a form of “infinity” and denote it using the last letter in the Greek alphabet. 1. 1} {0}
A definition for infinity When we write out any particular natural number p in the form of a set of smaller numbers. omega. in lowercase (ω). 4} {0. There are infinitely many infinite ordinals! We won’t delve into their properties here. p − 1} which has p elements ↓ and so on. 3.. This set. 1. according to the scheme we’ve just invented for building natural numbers? Can the set N represent any number at all? We can never finish writing the list of its elements. that set has exactly p elements in it:
0 = ∅ which has no elements 1 = {0} which has one element 2 = {0.} Think about this set N for a minute. We can imagine it. “Omega” is a traditional expression for “the end of all things. 1. 2} which has three elements ↓ p = {0.How Natural Numbers are Made
37
Figure 3-3
The natural numbers can be built up.. 3.

{∅.38 Natural Numbers and Integers
about them. 1} = {∅. postal zone of residence. But this method doesn’t increase the number of elements in each set as the defined number gets bigger. 1. {∅}. To make a good definition of the natural numbers. just as you can classify people according to blood type. {∅}}. It helps if you write the numbers as sets of smaller numbers and then break those expressions down: 0=∅ 1 = {0} = {∅} 2 = {0. forever Why can’t we do that to construct the natural numbers? Well.”
Solution
Start with 0. showing how each number can be assembled from “multiple nothings. {∅}} 3 = {0. {∅}. perhaps. 2 in Chap. 2. Each of the above sets. there are plenty of Web sites you can explore. or even (maybe someday) planet of residence. 2.
Are you confused?
In the solution to Prob.
. {∅. 3} = {∅. except for the first. {∅}}} 4 = {0. {∅. {∅}}}}
Special Natural Numbers
You can classify natural numbers in various ways. Here are a few of the most well-known types of natural numbers. 2} = {∅. country of residence. which is equal to ∅. contains one element. we saw how sets can be built on each other by “tacking on braces” to the null set: ∅ {∅ } {{∅}} {{{∅}}} {{{{∅}}}} {{{{{∅}}}}} ↓ and so on.
Here’s a challenge!
Write down the natural numbers 0 through 4 purely in terms of braces and the null set symbol. {∅}. {∅. 1. Go to your favorite search engine and enter “infinite ordinal” or “transfinite ordinal” in the phrase-search mode. number theorists want to have the sets contain more and more elements as the defined numbers get larger. we could.

. If you subtract 1 from every number in the set Nodd and then divide the result by 2. 6/2. 1×2. But you’ll never see a natural number that is both even and odd.} = {(0×2)+1. This is the familiar set
Neven = {0. . If you divide every number in the set Neven by 2.. 10/2. 5. preferably a fairly large one. 4. 2. 10.} Now do the same thing. 2×2. 9. 3×2. (4×2)+1.} Now do the same thing. 5....} = {(1−1)/2. 4.. Let’s pick a natural number.} = {0/2. or 9. 5..}
Odd numbers An odd natural number is a whole number whose numeral ends in 1. as large as you want.} = {0×2.. 6.. . If you multiply every number in the set N by 2 and then add 1 to the result. 4. 2. you get the set Neven of all the even natural numbers. (5−1)/2. but backwards. This is obvious... 2.. you get the set N of all natural numbers: N = {0. 4/2. . 1. you get another natural number. you get the set N of all natural numbers: N = {0. 5. (11−1)/2 . (2×2)+1. 8. (7−1)/2. when one of the numbers is 0 or 1. 4. and it will always be either even or odd.} The union of the set of all the even natural numbers and the set of all the odd natural numbers is the entire set of natural numbers. (5×2)+1. 2. (3×2)+1. 3. 11. Any number times 0 is equal to 0.. 2/2. 4×2.
Factors Whenever you multiply two natural numbers together. . 3. (3−1)/2. How about 99? How can we break this number down into a product of other natural numbers besides itself and 1? It’s easy to see that 33 and 3 will work:
33 × 3 = 99
. . 1. When you start working with the other numbers. (1×2)+1. You might find this mouthful of words easier to understand if you write it in symbols: Neven ∪ Nodd = N This means that you can pick any natural number. and any number times 1 is equal to itself. 7. even trivial... or 8. . 3. 7. If you multiply every number in the set N by 2. you get the set Nodd of all the odd natural numbers:
Nodd = {1. 6.. 3. 5×2.Special Natural Numbers
39
Even numbers An even natural number is a whole number whose numeral ends in 0.. (9−1)/2. but backwards. things get more interesting. 8/2.

How about 33? This can break down into the product of 11 and 3: 33 = 11 × 3 We can’t break 11 down into a product of natural numbers other than itself and 1.
Order 1st 2nd 3rd 4th 5th 6th 7th 8th Prime 2 3 5 7 11 13 17 19 Order 9th 10th 11th 12th 13th 14th 15th 16th Prime 23 29 31 37 41 43 47 53 Order 17th 18th 19th 20th 21st 22nd 23rd 24th Prime 59 61 67 71 73 79 83 89
. If you ever get a whole-number quotient as you go through this process. The square root of a number is a smaller number. When you want to find the prime factors of a large natural number. The first 24 prime numbers. Then add 1 to that number. Call this new whole number s. you can get some help from a calculator that has a square root key. “chop off ” any nonzero digits that might appear after the decimal point.40 Natural Numbers and Integers
We can’t break the number 3 down into a product of natural numbers other than itself and 1. Here’s how the process goes. Any nonprime natural number can be factored into a product of two or more primes. so you get a whole number. If this doesn’t go high enough for you.
Prime and composite numbers An “unbreakable” natural number is called a prime number. that gives you the original number when multiplied by itself. First. there are plenty of lists of primes on the Internet. and the whole product is called a composite number. Now divide the original number by all the primes (referring to Table 3-1) less than or equal to s. Any natural number. The number 1 is not considered prime. not always whole. It’s a natural number larger than 1 that can only be factored into a product of itself and 1. can be factored into a product of primes. Now we have 99 as a product of “unbreakables”: 99 = 11 × 3 × 3 Whenever you have a natural number expressed as a product of other numbers. Once you have done that. starting with the largest prime and working your way down. or simply a prime. The numbers in such a product are called the prime factors. Table 3-1 lists the first 24 prime numbers. use a calculator to find the square root of the number you want to factor. then you know that the divisor and
Table 3-1. those other numbers are called factors. The process of breaking a number down into a product of other numbers is called factorization or factoring. no matter how large.

A good example is 39. Perfect squares. you can sometimes tell right away that it’s not prime. If you take the square root of a perfect square. but it turns out that they are not.Special Natural Numbers
41
the quotient are both factors of the original number. Once you’ve found all the primes smaller than s that divide your original number without giving you a remainder. Still another is 57. If its numeral ends in an even number. then it can be factored down further. the order agrees with the number squared. you know that one of its factors is 2. are easy to find if you have a calculator. Table 3-2 lists the first 24 perfect squares.
Table 3-2. It can be factored into 13 × 3. Don’t stop dividing the original by primes until you get all the way down to 2. Some product of these will give you that original number.673 = 11 × 3 × 3 × 3 × 3 × 3
Perfect squares Whenever you multiply a number by itself. three times. unlike primes. The first 24 perfect squares. When you take the square root of a perfect square. you get a perfect square. or more. Note that we start with the “0th” rather than the “1st” in order here. which can be factored into 17 × 3.
Are you confused?
If you see a large natural number. numbers seem at first as if they ought to be prime.
Order 0th 1st 2nd 3rd 4th 5th 6th 7th Square 0 1 4 9 16 25 36 49 Order 8th 9th 10th 11th 12th 13th 14th 15th Square 64 81 100 121 144 169 196 225 Order 16th 17th 18th 19th 20th 21st 22nd 23rd Square 256 289 324 361 400 441 484 529
. and sometimes it won’t be. you always get a natural number. Sometimes. the process is called squaring. Sometimes the quotient will be prime. Another is 51. which can be factored into 19 × 3. The numbers 0 and 1 are included here. You might find that one of the primes is a factor twice. and the product is called a square. That way. If you multiply a natural number by itself. you always get a natural number. If it isn’t prime. For example: 99 = 11 × 3 × 3 297 = 11 × 3 × 3 × 3 891 = 11 × 3 × 3 × 3 × 3 2. with 0 and 1 included. you have found the prime factors of your original number. so it can be factored into a pair of natural numbers other than 1 and itself.

4.
Here’s a challenge!
When an even number is multiplied by 7.
Solution
For the first few even natural numbers.
Natural Number Nontrivia
Here are some interesting facts about natural numbers. Remember how you arrange the numerals on the paper and then do the calculations. I was about to call them “trivia. 8 (the second digit in 28). then the last digit in the product must be 0. multiplication by 7 always gives you an even number. getting the last digit of the product. If the number on the bottom is 7. if you can’t factor it into primes). the result always even. then your number is prime. Now think of “long multiplication” by 7. 2. or 8. You always start out by multiplying the last digits of the two numbers together.e. 2 (the second digit in 42). If the only factors you get are itself and 1 (i. which you are multiplying by the number on the bottom.. such as the “divisibility” tricks you’ll see later in this chapter. Here are the examples for all the single-digit even numbers: 0×7=0 2 × 7 = 14 4 × 7 = 28 6 × 7 = 42 8 × 7 = 56 You can prove that multiplying any even number by 7 always gives you an even number if you realize that the last digit of an even number is always even. Show why this is true. 4 (the second digit in 14). must look like one of the following: ______0 ______2 ______4 ______6 ______8 where the long underscore represents any string of digits you want to put there.42 Natural Numbers and Integers
The best way to find out whether or not a large odd number is prime is to try to factor it into primes.” but after thinking about it for awhile. 6. This number p. however large it might be. I decided that the ones involving primes are not trivial at all!
. There are some other techniques you can use determine when a number is not prime. The even number on top. Think of an even number p—any even number. The product of any even number and 7 is therefore always even. must end in 0. or 6 (the second digit in 56) respectively.

“Is there a largest prime?” The answer is “No. you might ask. but if there is a largest prime. • A natural number is divisible by 9 without a remainder if the sum of the digits in its numeral is a natural-number multiple of 9. Now that we have decided there is a largest prime.
Is there a largest prime? Now that you know what a prime number is. that have ever been proven in mathematics. If you can accept each step of this argument one at a time.” to find out if a natural number is divisible by 7 without leaving a remainder. 5. and you know that any nonprime natural number can be broken down into a product of primes. we can eventually write all of the primes... because it is a product of primes. p} Suppose that we multiply all of these primes together. • A natural number is divisible by 6 without a remainder if it is even and the sum of its digits is a natural-number multiple of 3. we can list the entire set of prime numbers (call it P). but the following rules can be interesting anyway. 13. and then divide that number by 2 again and get an even number. 11. There aren’t any convenient tricks. How about p? Theoretically. or theorems.” Here’s why.Natural Number Nontrivia
43
Divisibility If you want to know whether or not a large number can be divided by a single-digit number without leaving a remainder. Then we’ll prove that this assumption cannot be true by “painting ourselves into a corner” where we end up with something ridiculous. We get a composite number. You can combine these tricks and get the following facts: • A natural number is divisible by 4 without a remainder you get an even number after dividing it by 2. It might take mountains of paper and centuries of time. You can use a calculator to see immediately whether or not any number is “cleanly” divisible by any other. other than using a calculator or performing “long division. 3. Try to follow it step-by-step. 7. • A natural number is divisible by 8 without a remainder if you can divide it by 2 and get an even number.
• A natural number is divisible by 2 without a remainder if it is even. We can describe the set P in shorthand like this:
P = {2. that’s good enough.. No doubt. this number is huge—larger than any calculator can
. • A natural number is divisible by 10 without a remainder if its numeral ends in 0. • A natural number is divisible by 3 without a remainder if the sum of the digits in its numeral is a natural-number multiple of 3. there are some handy little tricks you can use. . Let’s start by imagining that there actually is a largest prime number. You might have to read the following explanation two or three times to completely understand it. • A natural number is divisible by 5 without a remainder if its numeral ends in either 0 or 5. suppose we give it a name. The fact that there is no such thing as a largest prime is one of the most important facts.

This time it’s going to be easy. we get a remainder of 1. So z is composite. we always get a remainder of 1. but .615 = 3 × 3 × 3 × 5 × 7 × 7 You will get into trouble if you say. Let’s call it y. But every natural number larger than 1 is either prime or composite! But . you can express it like this: z=y+1 = (2 × 3 × 5 × 7 × 11 × 13 × .615 is
. “The set of prime factors of 6. you must be sure to include all the occurrences of a prime factor if it appears more than once.. Because z is composite. Take this example: 6. 2 × p or 3 × p or 5 × p or 7 × p. × p) + 1 Now we know that z has to be larger than p.. and z is exactly 1 more than y. we can list them all.
Are you confused?
When you have found the prime factors for a composite number. Let’s start out by assuming that the number of primes is finite. that there is a largest prime number.. 5.615 is {3. But it can’t be prime either. because we’ve already determined that z is bigger than p. must be false. getting a number even larger than the product of all the primes? If you call that new number z. But there’s something else interesting about z.” How do you know whether a given factor occurs once. and load up our reductio ad absurdum “cannon” for another shot. or more? Some people get around this issue by putting a little superscript called an exponent after a number in the set to indicate that it occurs more than once as a factor. you can write the product out in any order.. Contradiction! The number of primes can’t be finite. comes to the rescue again!
How many primes? The discovery that there is no largest prime number leads us straightaway into another important truth: there are infinitely many prime numbers.. When a mathematician proves a major theorem like the one we just explored. three times. 5 at the end of Chap. If we divide z by any prime number. that is.. 7}. but . that secondary fact is called a corollary.. say. But we just discovered that there is no largest prime. 2. because z is 1 more than. But unlike a listing of the elements of a set. one element of set P. z can’t be composite. We know that z can’t be prime. That means one of them has to be larger than all the others. Our original assumption. Therefore. If the number of primes is finite. so it must be infinite. They write that the set of prime factors of 6. in which you are allowed to list any element only once.. it must be divisible without a remainder by at least one prime. so it is the largest prime. But wait! We just figured out a minute ago that if we divide z by any element of P. Reductio ad absurdum. That’s because if we divide y by any prime. and then some other fact follows on its heels. and we have already assumed that p is the largest prime.44 Natural Numbers and Integers
display—but it will be finite. What if we add 1 to y. which we first encountered in the solution to Prob. we are trapped! There’s only one way out of this situation. there’s no remainder. twice.

72}. but it helps if you start with the smallest factor and go up. By elimination. We can factor 2 out of any such number. listing each factor as many times as it “deserves. etc. the result became known as the set of integers.) are prime. negative numbers weren’t taken seriously. −1. exactly? That question is deeper than it seems at first thought.The Integers
45
{33. 4. nearly everyone has a credit card.. These days. That’s okay if you can remember that 33 does not literally mean 27 in this context. 3. or start with the largest factor and go down. −3. where 32 degrees represents the freezing point of water.. You can arrange the product in any order. That means you haven’t put any money in the bank that gave you the card. but you don’t owe the bank any money. and people outside the United States. That set is symbolized Z. Then people start calling temperatures negative. 5.. it has a balance of 0. we are adding 1 to an odd number. temperatures often get colder than 0 degrees. How could you have less than none of anything? When the set of all negative whole numbers was finally joined together with the set of natural numbers.}
Negative numbers What is a negative number. We can start to answer it by creating situations where negative numbers are really useful. Are there any other examples of two consecutive whole numbers that are both prime?
Solution
No. and they are also consecutive whole numbers. 0. Here’s another real-life situation where negative numbers come in handy. then. where 0 degrees represents the freezing point of water.
Here’s a challenge!
The numbers 2 and 3 are both prime. 10. most nonscientific people use the Fahrenheit temperature scale. like this: Z = {. If we take any of these primes and then find the next natural number. 1. −2.. which we have just seen can’t be prime. What if you buy some items at the local department store. There is only one way to factor a composite number into a product of primes.e. When you first get the card. “charging” up a balance of $49? How much money is in the account now? If you think of it as the bank’s
. The conclusion: when we come across any prime number larger than 2. 2. . use the Celsius temperature scale. In the United States.. In either system. and we always get a natural number bigger than 1. 6. either. 8. the next consecutive natural number is always composite. That always produces an even number. and 72 does not literally mean 49. This fact is called the Fundamental Theorem of Arithmetic. none of the even numbers larger than 2 (i.. all the primes larger than 2 are odd. Scientists.
The Integers
Centuries ago. (Neither 27 nor 49 are prime!) The clearest way to express the prime factors of a composite number is to write out the product.” and using multiplication symbols between them.

) Negative whole numbers are denoted by putting a minus sign in front of a natural number. 1. 1. 1} – {0. then to −$59. you’ll end up with negative $59. the bank will put a limit on it. You have. 2} {0. If you go to another store and charge $10 more. The same thing can happen with temperature. 2}
Proceed forever!
Figure 3-4 The negative numbers can be built up from the positive ones
by inventing an imaginary “number reflector” that reverses the “sense” of every natural number and gives it a “twin. negative $49.”
. put minus signs in front of all the numbers on one of the rays. In theory.46 Natural Numbers and Integers
account. (In practice. where a negative sign doesn’t change the meaning. in a sense. The exception is 0. making sure to include 0 so we get the entire set of integers? We can take two natural-number rays (or half-lines). How can we add the negative natural numbers to the “normal” or positive ones. in dollars. and then stick the rays together end-to-end so “positive 0” and “negative 0” are on top of each other. Figure 3-4 shows how this works. there is no limit to how large negatively your account. it was −10 degrees in the morning. You might think of the
Proceed forever!
3 2 1 0 “Number reflector” –1 –2 –3
{0. If you think of it as your account. you’re $49 dollars in debt. can become. they have a claim to $49 of your money. In the credit-card situation just described. 1} {0}
– {0} – {0. you start out with $0 and then go to −$49.
A “number reflector” We’ve already shown how the natural numbers can be generated from sets. If it was 0 degrees yesterday afternoon and then the temperature fell by 10 degrees overnight. “Negative 0” is the same thing as “positive 0” in ordinary mathematics.

2 . Start at 0.2} {0. and keep in mind that what you’re about to read does not represent the only way the set of integers can be defined in a “pure” way. Put your “abstract thinking cap” on again (if it isn’t glued to your head by now). 3}
Figure 3-5 Here’s a way to generate the set of integers
with a scheme similar to the one we used to build up the set of natural numbers. suppose you hop alternately back and forth. 3-5. but in mathematical terms.” and their negatives as being attached to a ray that dangles straight down. positive. Take a look at Fig. using only the idea of a set and nothing else. negative. and then from 2 to 3. 1} {0. Then go down two units to −1. that’s the impression you’ll get if you look at Fig. always moving in the same direction. -1. -1.” In order to define the negative numbers this way. 1. Instead of hopping from 0 to 1. 1. and 0. but in the “mathematical cosmos” we have powers that ordinary mortals lack. it is a little “impure. -2 } {0. Then go up three units
3 2 1 Proceed forever! 0 -1 -2 -3
{0. -1} {0} Start here {0. But the integers go on forever in two directions. and cover every point on it? You have to pick one direction or the other. We can define the entire set of integers in the same way as we defined the set of natural numbers. A pure mathematician would demand some way to define all the integers. 1.
Building the integers The natural numbers have a clear starting point. 2 . How can you start moving along a line that goes in two directions. right? Wrong! In the real world that might be true. 1. then move up one unit to 1. which is 0.The Integers
47
natural numbers as being attached to a ray that stands straight up above the “number reflector. This is a fine way to imagine the integers. and then from 1 to 2. we have to come up with new gimmicks that we did not need to define the natural numbers. . 3-4. -1. 2 .
. At least.

3-5. you can go to an old-fashioned chalk blackboard (every good mathematics party has one.48 Natural Numbers and Integers
to 2. −13. But −158 is larger negatively than −12 or −32 or −157. Keep hopping alternately up and down. {∅}}. It is therefore a product of primes that are all positive. and any negative integer is smaller (or less) than any natural number. right?) and scribble out the following to make your point: 0=∅ 1 = {0} = {∅} −1 = {0. −1.. really?” you can say. −2 is larger (or greater) than −5. we get the negative of that composite number. down six units to −3. {∅. the integers themselves are shown to the left side of the vertical line. making your hop one unit longer every time. 1. Conversely. 3-4 or 3-5. −5. isn’t it a bigger problem than if you are in debt by $12? In the literal sense. To avoid confusion when comparing numbers. we can define it as the set containing 0. {∅}. −3. as big or small as you want. You’ll eventually reach it if you make enough hops. {∅}} 2 = {0. −11. up five units to 3. −1. and any natural number is larger (or greater) than any negative integer. does that make all the nonprime negative numbers composite?
Solution
Let’s keep the traditional definition of composite number: a product of two or more primes. just as −158° is colder than −12°. −19. But if you like. and 2. 1. and larger as you go upward. If you draw a number line and represent the integers as points on it.{∅. {∅}. Pick any integer. 2} = {∅. The next time you are at a party with a bunch of mathematics lovers and somebody asks you. and so on. the best policy is to be careful with your choice of words. −2. If we make one of those primes negative. But that can begin to seem strange if you think about it awhile. −1} = {∅. −5 is smaller (or less) than −2. {∅. In Fig.
Here’s a challenge!
If we allow all negatives of primes (i. −7. {∅. what does it mean if one number is “larger” or “smaller” than another? How about the expressions “less than” or “greater than”? A mathematician will tell you that the integers get smaller as you move downward in Figs. 1. For example. If you want to bring down the house. forever
Are you confused?
The integers can get confusing when you compare values. {∅}}}} ↓ and so on. Now imagine that we have some positive composite number. the integer −158 is less than −12 or −32 or −157. built up as sets of previously defined integers. −158 is indeed smaller (or less) than −12. that can be debated. How can −158 be “smaller” than −12? If you find yourself in debt by $158. “What is the number −2.e. are shown on the right side. and their equivalents. positive or negative. {∅}.{∅}}} −2 = {0. Figure 3-6 should clear up any lingering uncertainty you might have about this. down four units to −2. 3-4 or 3-5.” That should get you a raised eyebrow. For example: 100 = 5 × 5 × 2 × 2
. −17. such as is done in Figs. “Well. In fact. 1} = {∅. …) to be called prime.

Practice Exercises
49
Larger positively Smaller positively
3 2 1 0 -1 -2 Smaller according to the traditional definition Larger according to the traditional definition
Smaller negatively
Larger negatively . we can see that −100 = −5 × 5 × 2 × 2 This same technique can be applied to any negative nonprime number smaller than −3 to show that it’s composite! We have to be sure that “negative primes” are allowed in the mathematical system we’re dealing with.3
Figure 3-6 This drawing.
If we remember the basic multiplication sign rules.
. can help you
avoid confusion when comparing the values of integers. If not. and careful choice of words. The number 3 is odd. According to the traditional definition. Don’t hurry! You’ll find worked-out answers in App. so this trick won’t work. When an odd number is multiplied by 3. A. show a counterexample (a situation where an odd number is multiplied by 3 to get an even number). You may (and should) refer to the text as you solve these problems. all the primes are natural numbers larger than 1. demonstrate why.
Practice Exercises
This is an open-book quiz. is the result always odd? If so. The solutions in the appendix may not represent the only way a problem can be figured out. then what number does nothing represent? 2. If the number 0 is the set containing nothing. by all means try it! 1. does that mean n must be odd? 3. If you think you can solve a particular problem in a quicker or better way than you see there. If a number n is divisible by 3 without a remainder.

3-5 with its pattern of dashed. What interesting property does this number have? 7. Find out whether or not 901 is a prime number. What interesting property does this number have? 8.197 into a product of primes. Here’s a hint: Use Fig. arrowed guidelines to create an “implied one-ended list” of the integers that captures them all. Break down 2. Can you think of a good reason why the natural numbers 0 and 1 are not defined as prime? Here’s a hint: It should never be possible for a number to be both prime and composite. 5.081 into a product of primes. Are any negative integers composite if we insist on using the traditional definition of a prime number? 9. 10. Break down 841 into a product of primes. Break down 1.
. Show how the natural numbers can be paired off one-to-one with the integers.50 Natural Numbers and Integers
4. 6.

Think of upward distances as positive displacements. When you see something like a + b = c.
. That gets us to the point for −3 + 2. As an example. because we’ve already illustrated the number line in a vertical sense.
Meet the variables! When you want to talk about how numbers relate to each other but don’t want to specify any particular numbers. The direction (up or down. The term variable means that a quantity doesn’t have any fixed value. but only that they are related in a certain way. For variables representing integers.52 Addition and Subtraction
3 2 |2| = 2 1 0 -1 |-3| = 3 -2 -3
Figure 4-1 The absolute value of a number is its distance from 0 along
the number line. you know you are supposed to add a quantity a to another quantity b to get a third quantity c. 4-2. It happens to be −1. you can use variables instead. we first find the point on the number line representing a. We start at the point for −3 and move up 2 units. You don’t have to know what the actual numbers are. and downward distances as negative displacements. Then we move up b units. mathematicians most often use small. positive or negative) doesn’t matter. If we have an integer a and we want to add another integer b to it. it can vary. To add. as shown on the left side of Fig. We’ll go up and down here. you can get the numeral representing its absolute value by removing the minus sign.
you have a numeral that represents a negative integer. italic letters from a through q. That is why absolute values can never be negative. That will get us to the point representing a + b. suppose a = −3 and b = 2. move upward Now let’s get back to displacement.

What if we reverse the order of this sum? We start with 2 and travel upward −3 units. you’re right. so negatives can make sense! An upward displacement of −3 units is the same as a downward displacement of 3 units.Moving Up and Down
53
3 2 1 0 Finish here Move upward by 2 units Start here -1 -2 -3 Finish here Start here
Move upward by -3 units
Figure 4-2 On the left. We’re talking about displacement here. we add −3 + 2. You see that a rain shower has caused an existing jam in the westbound traffic on Boxelder Bug
. When we move negatively upward. 4-2. Suppose you fly a rush-hour traffic observation helicopter for your local TV station. move downward If you think it’s ridiculous to imagine downward movement as negative upward movement. which corresponds to −1. and it won’t seem to make sense. we end up at the same point. except for one little catch. not simple distance. You are going to come up with situations in mathematics where you’ll get a negative quantity for an answer to a problem. On the right. When we add the integers −3 and 2 in either order as shown. We have now analyzed these two facts: −3 + 2 = −1 and 2 + (−3) = −1
To subtract. we move downward by the equivalent distance. This process is shown on the right in Fig.
we add 2 + (−3).

”) If you think about this for a little while. The jam has moved in the opposite direction from the flow of traffic! Look again at the right-hand side of Fig. That will get you to the point representing a − b. Adding a negative number is the same thing as subtracting the absolute value of that number. You finish at the point corresponding to 2.” and not “a huge gathering of people. you get 2.54 Addition and Subtraction
Boulevard to be displaced by −4 miles. You add a negative number to some other quantity. Write it down in the simplest possible form. “convention” means “custom” or “way of doing things. The answer is that there’s no need for a plus sign in a positive number. Figure 4-3 is a number-line drawing that shows how this works. 4-2. you move negatively downward. first find the point on the number line representing a. It is a mathematical convention. but positive numbers don’t have a plus sign? Is there something technically wrong with including a plus sign so people know when a number is positive? Why leave any doubt? That’s a good question. which means you really move up 5 units. this fact is written −3 + 5 = 2
. Using variables. meaning that you actually travel upward. Why write +3 + (+7) when you can write 3 + 7?
Here’s a challenge!
In terms of the integer line. You start at −3 and move down −5 units. A number is always assumed to be positive unless there’s a minus sign in front of it to indicate that it’s negative. it makes sense. you can write that statement as a + (−b) = a − |−b| which means the same thing as a + (−b) = a − b If you have an integer a and you want to subtract another integer b from it.
Are you confused?
Have you been wondering why negative numbers always have a minus sign in front of them. express the fact that when you subtract −5 from −3. You can write −3 − (−5) = 2 In its simplest possible form. Then travel down b units.
Solution
When you subtract a negative number. (In this context. This means the jam has been displaced to the east by 4 miles.

Identity. any integer. That integer is 0. that govern addition and subtraction.
3 2 1 0 -1 -2 -3 Start here Move downward by -5 units Finish here
Identity. we go upward by the equivalent distance. The word “identity” means that you always get an output value identical to the input value.
The identity element There is one special integer that you can add to. When we go negatively downward.” although that term is informal. and Signs
Let’s review the mechanics of addition and subtraction. These laws apply not only to the integers. and numbers you’ll encounter in algebra. Because adding 0 to a number gives you the same number again. Grouping. For any integer a. and make sure you understand how the signs work. You can also call it the “subtractive identity element. we start with
−3 and then subtract −5. Grouping. expressions. called laws. the following two statements are always true:
a+0=0 and a−0=0
. Then we’ll proceed to two important rules. and it will never change the value. or subtract from. 0 is called the additive identity element. ending up with 2. but to all quantities. and Signs
55
Figure 4-3 Here.

“I drove 200 miles less last month!” Then it comes to you. Always be sure the number of opening parentheses in an expression is the same as the number of closing parentheses. When you subtract a positive. the result grows smaller. the result grows smaller. it’s like subtracting a positive.56 Addition and Subtraction
Grouping with parentheses A couple of minutes ago. not 2. William says. you saw some expressions containing parentheses around a negative number or quantity. When you write an expression. be sure to include parentheses when you need them. Should you subtract 5 from −3 twice? That would give you −13.” Anna says. It’s a problem because it’s ambiguous.” Maria says. One after another. “I drove 45 fewer miles. People go to meetings once a month to share information about how much less they’ve driven in the past month compared with the month before that. the result grows larger. Does the double minus sign mean something different from plain old subtraction or addition? Pay attention to parentheses when you see them. For any two integers a and b. When you subtract a negative. “I cut back by 65 miles. You have
. Here’s one of those expressions. the members of the group tell their stories.
Signs in addition and subtraction Here’s a summary of how addition and subtraction work for negative integers as well as for positive ones:
• • • • When you add a positive. the result grows larger. a + (−b) = a − b and a − (−b) = a + b
Are you confused?
Here’s a real-life situation where the idea of subtracting a negative number makes pretty good sense. When you add a negative.” If you don’t write the parentheses. with the right-hand side changed to a variable:
−3 − (−5) = a These parentheses indicate that you should consider −5 as “negative 5. it’s like adding a positive. Suppose the people in your town are trying to reduce their driving because of high gasoline prices.
You might want to memorize two general facts. the above expression is −3 − −5 = a This is an example of improper mathematical grammar.” not “the subtraction of 5. People can’t be sure what it means. You can add extra parentheses to an expression as long as you don’t change the meaning. Should you subtract 5 from −3? That would give you −8. They’ve started a support group.

”
Here’s a challenge!
Start with the integer 5. It means you can commute (interchange) the two numbers you’re adding. negative. In formal terms. then again from bottom to top. Here we go: 5 + (−3) = 5 − 3 = 2 2 − (−6) = 2 + 6 = 8 8 − 10 = −2 −2 + 14 = 12 12 − (−21) = 12 + 21 = 33
The Commutative Law for Addition
In basic arithmetic. you learned that you can add a long string of numbers backward or forward. clear your throat. just to be sure they have done the arithmetic correctly. then subtract −6 from that.
It fails when you subtract In subtraction. The group leader asks. Good accountants take advantage of this when checking their work. or more numbers in a sum.The Commutative Law for Addition
57
not done so well.
a+b=b+a This works whether the numbers are positive. and get the same sum either way. the order does matter. and finally subtract −21 from that. It’s easy to find an example that shows why. as long as the number of addends is not infinite.
It works when you add The fact that you can add two integers in either order and get the same result is called the commutative law for addition. called the addends. a mathematician would say that for any two integers a and b. “How much did you cut back? Go on. or 0. and declare. Consider this:
3 − 5 = −2
. It also works if there are three. They’ll add up a column of numbers from top to bottom. then add 14 to that. add −3 to it. and it doesn’t matter.” You blush. then subtract 10 from that. five. four. “I reduced my driving last month by negative 80 miles. hold your head up. don’t be afraid to tell us. paying careful attention to signs and using parentheses when we need them. What’s the result?
Solution
We can break this down step-by-step.

but you might find it useful as a memory aid.” For example. and you’ll always end up with the same final balance—if you don’t make any calculation errors!
. a check for $25 becomes a “negative deposit” of −$25. we can apply the commutative law and get a + (−b) = −b + a Now we can combine the above two equations into a three-way equation: a − b = a + (−b) = −b + a Then we can get rid of the middle term and write a − b = −b + a Let’s call this the “inside-out commutative law for subtraction. Consider that as a “starting deposit.
Are you confused?
Imagine that you have a checking account and your balance on January 1 was exactly $700. Now you can add all the “positive deposits” and “negative deposits” in any order. Remember that adding a negative is the same as subtracting a positive. It only works if a and b happen to be the same.
Turning it inside-out We can use a trick that will make the commutative law “sort of work” with subtraction. This trick involves taking every subtraction and turning it into the addition of a negative number. it is almost never true. You convert all the checks to “negative deposits. it is not always true that a−b=b−a In fact.58 Addition and Subtraction
but 5−3=2 In formal terms we would say that for any two integers a and b. for any two integers a and b.
a − b = a + (−b) Because the right-hand side of this equation is an addition problem. That is.” By the end of June. You want to figure out your balance as of June 30. you’ve made 15 deposits and written 20 checks.” That’s not a formal name. This trick is often used by accountants who must work with long columns of credits (money added) mixed with debits (money taken away).

You drive 25 miles north. The simplest case of this rule. called the associative law for addition.” Then you your trip is a sum of displacements like this: −25 + 45 + (−50) + 7 + (−12) + 49 + 5 = 19 Again. and in this order: 25.
Solution
First.The Associative Law for Addition
59
Here’s a challenge!
Suppose you have bought a new car and you want to go for a test drive. Now imagine that driving south is “positive mileage” and driving north is “negative mileage. b. consider driving north as “positive mileage” and driving south as “negative mileage. that’s 19 miles south. and it what direction. 12. 50. then turn around and drive 45 miles south. −7. You can lump the addends together any way you want.” Then you drive the following distances in miles. and finally 5 more miles south. 12 miles north.
The Associative Law for Addition
Another major rule that applies to addition involves how the addends are grouped when you have three or more numbers. will you finish? Solve this problem in two different ways.
It works when you add Here’s how a mathematician would state the associative law for three addends. −5 Add these all up: 25 + (−45) + 50 + (−7) + 12 + (−49) + (−5) = −19 That’s 19 miles south of your home town. You live on a flat plain that seems to stretch forever in all directions. 49 miles south. Then you turn around again. −45. Then you go 7 miles south. involves sums of three integers. For any three integers a. You start driving on a straight highway that runs north and south for hundreds of miles on either side of your home town. and you’ll always end up with the same result. and c
(a + b) + c = a + (b + c) For instance: (3 + 5) + 7 = 8 + 7 = 15 and 3 + (5 + 7) = 3 + 12 = 15
. How far from your home town. −49. driving 50 miles north.

we would say that for any three integers a. the grouping does matter.
Mixing the signs How about mixed addition and subtraction? Let’s try it with some integers. It’s easy to see why you can’t apply the associative law to subtraction and get away with it. Proving that a law always works is more difficult. or 0. You only have to find one case where it fails. It’s easy to prove that a law does not hold in every possible case.60 Addition and Subtraction
This works whether the numbers are positive. b. You have to use airtight logic. something has to work all the time. it fails! Now we know that with mixed signs. but not always. That’s what proofs are all about. You can’t do it using specific integers. called a counterexample. negative. because you’d have to try an infinite number of cases one at a time. “sometimes” does not suffice. the associative law sometimes works. Even “almost always” is not good enough. it is not always true that (a − b) − c = a − (b − c) The associative law hardly ever works with subtraction. and c.
. If something is to be called a law in mathematics. Consider this:
(3 − 5) − 7 = −2 − 7 = −9 but 3 − (5 − 7) = 3 − (−2) =3+2=5 In formal terms. What happens if we switch the positions of the signs? (3 − 5) + 7 = −2 + 7 = 5 but 3 − (5 + 7) = 3 − 12 = −9 This time.
It fails when you subtract In subtraction. In order to be a law.
(3 + 5) − 7 = 8 − 7 = 1 and 3 + (5 − 7) = 3 + (−2) = 1 It works in this case. It also works if there are four or more numbers in a sum. “Usually” won’t do the job either. to show that something can’t be called a law.

The Associative Law for Addition
61
Add. then subtract The associative law can work indirectly when subtraction is involved. (a − b) + c = a + (c − b)
Subtract. This looks messy. but a little junk is easy to tolerate when you realize that we’ve just proved something significant. We’ve shown that you can always use the associative law with mixed signs when the first operation is addition and the second one is subtraction. to indicate a grouping with another grouping inside. called brackets . then add Now let’s look the general situation where we have seen that direct application of the associative law does not always work. like this:
(a − b) + c = [a + (−b)] + c We can use the associative law for addition to get a + [(−b) + c] Now we can use the commutative law for addition inside the brackets to get a + [c + (−b)] We can simplify this to a + (c − b) Now we know that when the first operation is subtraction and the second is addition. then subtract again Finally. b. We can rearrange an expression like that as follows:
(a − b) − c = [a + (−b)] + (−c)
.
Subtract. if you change every subtraction into addition. so it becomes all addition. let’s explore the situation where we subtract twice. How can we modify it to make it work? We can rewrite an expression where the first operation is subtraction and the second one is addition. and c
(a + b) − c = (a + b) + (−c) = a + [b + (−c)] = a + (b − c) We use square parentheses. For any three integers a. It’s the same trick as with the commutative law.

that’s good enough. multiply by −1. That will come in the next chapter. What are you actually doing when you change c to −c ? Here’s an alternative to the “number reflector” idea. But you’ve had basic multiplication in your arithmetic classes. When you want to find the negative (also called its additive inverse) of any integer. This is a simple statements/reasons (S/R) proof. Because these statements are not very complicated and the proof is not too hard.62 Addition and Subtraction
We can use the associative law for addition to get a + [(−b) + (−c)] We can simplify this to a + (−b − c) Now we know that when both operations are subtraction. you can write it as a table with statements on the left and reasons on the right. It works like this: c × (−1) = −c and −c × (−1) = c
Here’s a challenge!
Based on the commutative law for the sum of two integers and the associative law for the sum of three integers. so let’s “cheat” for a moment and take advantage of that. Table 4-1 shows how it’s done. and c a+b+c=c+b+a
Solution
If you can manipulate the left-hand side of this equation to get the expression on the right-hand side. (a − b) − c = a + (−b − c)
Are you confused?
We haven’t started to dissect the anatomy of multiplication yet. show that for any three integers a.
. b.

on the average. Here is a proof that shows how you can reverse the order in which three integers a. If you think you can solve a particular problem in a quicker or better way than you see there. In long strings of sums and differences. evaluate the following two expressions. You want to find out if this is true. Using the rules explained in the previous exercise. Don’t hurry! You’ll find worked-out answers in App. The solutions in the appendix may not represent the only way a problem can be figured out.D.Practice Exercises
63
Table 4-1.E. Suppose someone tells you that there was a significant trend in the mid-winter average temperatures in the town of Hoodopolis during the period 1998 through 2005. translated into English as “Which was to be proved”
Practice Exercises
This is an open-book quiz. how should you evaluate the string of sums and differences if there are no parentheses at all? Here it is: 3 + 5 − 7 + 9 − 11 + 13 − 15 4. b. and then perform the operations outside the parentheses from left to right. each statement is equal to all the statements above it. Reasons Begin here Group the second two integers Commutative law for the sum of b and c Commutative law for the sum of a and (c + b) Ungroup the first two integers Latin Quod erat demonstradum. and c are added. To illustrate the importance of the placement of parentheses in a mixed sum and difference.
. by all means try it! 1. you should first perform the operations inside the parentheses from left to right. Evaluate and compare these two sums: a = |−3 + 4 + (−5) + 6| and b = |−3| + |4| + |−5| + |6| What general fact can you deduce from the results? 2. You may (and should) refer to the text as you solve these problems. You come across some old heating bills from the utility company that show how much warmer or cooler a given month was. As you read down the left-hand column. Here are the expressions: (3 + 5) − (7 + 9) − (11 + 13) − 15 and 3 + (5 − 7) + (9 − 11) + (13 − 15) 3. and get the same sum.
Statements a+b+c a + (b + c) a + (c + b) (c + b) + a c+b+a Q. A.

9. January 2000 averaged 6 degrees cooler than January 1999. Show at least one situation where can you say that a−b=b−a Don’t use the trivial case where a and b are both equal to 0.64 Addition and Subtraction
compared to the same month in the previous year. b. Don’t use the trivial case where a. and c are integers. Simplify and compare these expressions: (a + b − c) + (a − b + c) and a + (b − c) + (a − b) + c
. Based on the associative law for the sum of three integers. Based on the commutative law for the sum of two integers and the associative law for the sum of three integers. 7. b. Don’t use the S/R table method. c. 10. and call it by another name. b. You look at the records for January and find out that in Hoodopolis. Don’t use the trivial case where a. January 2001 averaged the same temperature as January 2000. January 1999 averaged 3 degrees warmer than January 1998. is called a lemma. January 2002 averaged 7 degrees warmer than January 2001. and c are all 0. Show at least one situation where you can say that (a − b) + c = a − (b + c) where a. b. b. when used to prove something new.
What was the difference in the average temperature between January 2005 and January 1998 in the town of Hoodopolis? 5. and c are integers. 8. January 2003 averaged 1 degree cooler than January 2002. January 2004 averaged 2 degrees warmer than January 2003. and c are all 0. and d a+b+c+d=d+c+b+a Here’s a hint: use the solution to the last “challenge” problem in this chapter as a shortcut. and d (a + b + c) + d = a + (b + c + d ) Do this in narrative form. construct an S/R proof showing that for any four integers a. c. 6. A previously proved fact. b. Show at least one situation where you say that (a − b) − c = a − (b − c) where a. • • • • • • • January 2005 averaged 5 degrees cooler than January 2004. prove that for any four integers a. Here’s a hint: “zip up” the sum b + c.

look at Figs. you add p to itself (n − 1) times. For examples. From the above facts. In a multiplication problem.
To divide.66 Multiplication and Division
12 8 Finish here 4 Start here Multiply by 3 Add to itself 2 times Get 3 times as far from 0 Start here Multiply by 5 Add to itself 4 times Get 5 times as far from 0 –4 –8 Finish here –12
Figure 5-1 On top. Figure 5-2 shows how this works for 2 × (−3) and −2 × (−5).
−2 is multiplied by 5. On the bottom. and then move away from the “number reflector” |n | − 1 times by a distance equal to the absolute value of p. so you end up
. To avoid clutter. you move away from the “number reflector” (n − 1) times by a distance equal to the absolute value of p.” you move up. 5-1 and 5-2 and think backward! In Fig.
Suppose p is an integer and n is a positive integer. first take the additive inverse (negative) of your starting integer p. you get a product. they are both called factors. only the even-integer points are shown on the number line. More often. 2 is multiplied by 3. the first quantity (the one to be multiplied) is sometimes called the multiplicand. The “number reflector” is shown as a horizontal. When you multiply two quantities. suppose you start with 6 and you want to divide it by 3. If your starting integer p is above the “number reflector. and the second quantity (the one you are multiplying by) is sometimes called the multiplier. On the number line. if your starting integer p is below the “number reflector. You reduce your distance from the “number reflector” by a factor of 3 but stay on the same side.” you move down. 5-1. move in Now imagine the multiplication process in reverse. What if n is negative instead of positive? To multiply p by n in this situation. dashed line. you can see that whenever you multiply p by n. Figure 5-1 illustrates how this works for 2 × 3 and −2 × 5.

If you start with 10 and divide by −5.” What must you multiply 0 by if
.
Division by 0 What happens if you try to divide an integer by 0? You don’t get any integer. You can add. the integers divide each other “cleanly” without remainders. In math jargon. Therefore. so you end up at −2. 2 is multiplied by −3. “Messy division” produces fractions. When you do a division problem with integers. the operation of division is not closed over the set of integers. or any other known type of number. Sometimes it is called a ratio. In Fig. Dividing by a negative number is trickier. Another example is the division of −7 by −12. You jump to the other side of the “number reflector” and then reduce your distance from it by a factor of 3. finishing at −2. only the even-integer points are shown on the number line. In a division problem.
at the point representing 2. 5-2.Moving Out and In
67
12 Finish here 8 4 Start here Multiply by -5 Take additive inverse Add to itself 4 times Get 5 times as far from 0 Start here -4 Finish here Multiply by -3 Take additive inverse Add to itself 2 times Get 3 times as far from 0 -8 -12
Figure 5-2 At lower left. subtract. Remainders occur when one nonzero integer divides another integer and the result is not an integer. you get a quotient. subtraction. the operations of addition. At upper right.
−2 is multiplied by −5. If you start with −10 and divide by 5. When you divide a quantity by another quantity. To avoid clutter. and the second number is sometimes called the divisor. you don’t always get another integer. you jump to the other side of the “number reflector” and then cut your distance from it by a factor of 5. and multiplication are closed over the set of integers. 6. finishing at 2. suppose you start with −6 and divide it by −3. you cut your distance from the “number reflector” by a factor of 5 and stay on the same side. In these situations. Look at this problem “inside-out. or multiply any two integers and always end up with another integer. the first number is sometimes called the dividend. which we’ll study in Chap. An example is the division of 5 by 3.

as you’ll see in Chap. no known number solves this problem. No matter how large you make n.68 Multiplication and Division
you want to get. That way. enduring definition of “infinity” produced by mathematicians has ever had anything to do with division by 0. by −2. “Division by 0 is not defined. Then the result becomes a new multiplicand. we need proof before we can claim something is true!
Manipulating equations Whenever we add or subtract a certain quantity to or from both sides of the equation. But I came pretty close. show what happens when you multiply the integer −1 over and over. you won’t get confused later on when we do something like divide both sides of an equation by 999. In fact. I never came up with a well-defined way to do it. you always get 0 when you add it to itself (n − 1) times. “Let’s define it.
Are you confused?
Have you heard that dividing a positive integer by 0 gives you “infinity”? If not. 11. we still have a valid equation. or multiply by can be a number. Because the multiplier is negative. you probably will some day. but that doesn’t prove that it will. a variable. endlessly. then!” She would repeat herself. The same is true if we multiply both sides of an equation by a certain quantity. “Why not?” or retort. In mathematics. If you divide both sides of an equation by a variable or an expression containing a variable. Because the
. I would ask. and had a lot of fun trying.
Solution
Figure 5-3 illustrates this process. 3? How many times must you add 0 to itself to get anything but 0? No integer can do this trick. you must add 0 to itself (n − 1) times. it’s best to stick to nonzero numbers.
Here’s a challenge!
In terms of the number line and displacements. If you divide both sides of an equation by anything. Be skeptical! The first thing you must do to figure out if it’s really true is to define “infinity. or divide either side by a certain quantity other than 0. Again. Most mathematicians will tell you that division by 0 is “not defined. as long as it is the same for the left-hand side of the equation as for the righthand side. we jump to the opposite side of the “number reflector” each time we multiply. No meaningful. or multiply both sides by (a + b).” That’s not easy.” I did not take her seriously. and I pestered her about it. look at the problem “inside-out. Why should adding 0 to itself forever make any difference? It’s tempting to suppose that it might. or a complicated expression.” That’s what my 7th-grade math teacher kept saying. subtract. you can get into trouble. say. Keep these rules in mind. The fact that we can do these things makes solving equations and proving various facts far easier than they would be otherwise.” If you want to multiply 0 by any positive integer n. so I started trying to make division by 0 work. The quantity you add.

we double our distance from the “number reflector” with each jump. We jump back and forth across the “number reflector.” doubling our distance from it with each jump. To avoid clutter. only the even-integer points are shown on the number line. forever
. Expressed as equations. we have −1 × (−2) = 2 2 × (−2) = −4 −4 × (−2) = 8 8 × (−2) = −16 −16 × (−2) = 32 ↓ and so on.Moving Out and In
69
To 32 8
2
Start at -1 -4
-16
Figure 5-3 Here is what happens when we start at
−1 and multiply by −2 over and over.
absolute value of the multiplier is 2.

In arithmetic. When you see −4/a. it means 8 divided by 2. it means a times b times c. When you see −4a.70 Multiplication and Division
Identity. When you see a/(b − c). Parentheses are placed around complicated expressions when they are multiplied by each other. Grouping. you can use the familiar “times sign” and put the numerals for the factors on either side. When you see (a + b)(c + d ). addition. we’ll use the forward slash (/) to indicate division.” Another symbol you’ll often see is the small. elevated dot (·). it means a divided by −4. 1 is called the multiplicative identity element. Then we’ll proceed to the three major laws that govern the interplay between multiplication. This symbol (×) looks like a tilted cross or a letter “x. When you see a(b − c). but that’s rarely used in algebra. it means 3 times 7. and subtraction.
• • • • • • • When you see 3 × 7.” but technically this term is okay. When you see a/b. it means (a + b) times (c + d ). When expressions are complicated. As with multiplication. it means a times (b − c). When you see a/(−4). the dividend (the number you want to divide) can be placed on top of a long horizontal line.) For any integer a
a1 = a
. When you see (a + b)/(c + d ). it means −4 times a. (For some reason I’ve never heard it called the “divisive identity element. parentheses are placed around complicated expressions when they are divided by each other. When a number is multiplied by a variable. it means (a + b) divided by (c + d ). it means a times b. it means a divided by b. or when a variable is multiplied by another variable.
The identity element You can multiply or divide any integer by 1. and it won’t change the value. and the divisor (the number you divide by) is placed underneath. Because multiplying or dividing by 1 always gives you the same number again. When you see −3·7.
Notation for multiplication When you want to multiply two numbers. When you see abc. and Signs
Let’s review how signs work in multiplication and division.
Notation for division In this book. it means −3 times 7. division.
• • • • • • When you see 8/2. it means −4 divided by a. it means a divided by (b − c). you’ll see their symbols run together without any space between. you sometimes see the dash with two dots (÷). When you see ab.

you reverse the sign but do not change the absolute value. Grouping. and Signs
71
and a/1 = a
The sign-changing element When you multiply or divide any integer by −1. but they would confuse some people. For any integer a
a(−1) = −a and a/(−1) = −a Conversely. Just be sure that for every opening parenthesis you put in. and a negative integer becomes positive.Identity. the above expressions are 3 × −5 = a and 15/−3 = b Expressions like these might be clear enough to you. Don’t be afraid to add parentheses to an expression if you think they will prevent ambiguity. A positive integer becomes negative.
Parentheses in simple products and quotients Look at these expressions:
3 × (−5) = a and 15/(−3) = b If you don’t write the parentheses.
. −a(−1) = a and −a/(−1) = a Note that a(−1) here means a times −1. but the distance from 0 on the integer line stays the same. you include a corresponding closing parenthesis later in the expression. Those parentheses are important! The integer −1 can be called the multiplicative sign-changing element or the divisive sign-changing element. not a minus 1.

0. but with no parentheses. as my dad would always say. • When you divide a negative integer by −2 or less. the result stays positive and the absolute value decreases (it gets closer to 0). • When you multiply a positive integer by −2 or less. Study these rules carefully. the result becomes positive and the absolute value increases (it gets farther from 0). Here’s an example. and division all mixed up. “plug in” some actual numbers and test them. subtraction. the result becomes negative and the absolute value decreases (it gets closer to 0). or moving upward on the integer line. the result stays negative and the absolute value increases (it gets farther from 0).
• When you multiply a positive integer by 2 or more. If any of them confuse you. • When you divide a positive integer by 2 or more. For any two integers a and b
a × (−b) = −ab = −(ab) and a/(−b) = −a/b = −(a/b)
Are you confused?
Suppose you see an expression with addition.72 Multiplication and Division
Signs in multiplication and division Here is a set of rules for multiplication and division by any integer except −1. or moving down. “homogenize”) two general facts. the result stays negative and the absolute value decreases (it gets closer to 0). the result becomes positive and the absolute value decreases (it gets closer to 0).
“Homogenize” these! You would do well to memorize (or. 2 + 48/4 × 6 − 2 × 5 + 12/2 × 2 − 5 You’ll get an answer that depends on which operations you do first. the result becomes negative and the absolute value increases (it gets farther from 0). multiplication. or 1. and “less” always means more negative. • When you divide a negative integer by 2 or more. • When you multiply a negative integer by −2 or less. In this order: • Group all the multiplications • Do all the multiplications from left to right
. Remember that “more” always means more positive. • When you divide a positive integer by −2 or less. the result stays positive and the absolute value increases (it gets farther from 0). You must use certain rules of precedence. • When you multiply a negative integer by 2 or more. Do not approach a mixed-operation problem like this by simply grinding out the arithmetic from left to right.

But you can’t expect to do the same thing if there is division anywhere in the process. then divide that result by −20. you learned that you can multiply numbers in any order and always get the same product.
. multiplication. subtraction.
Statements 2 + 48 / 4 × 6 − 2 × 5 + 12 / 2 × 2 − 5 2 + 48 / (4 × 6) − (2 × 5) + 12 / (2 × 2) − 5 2 + 48 / 24 − 10 + 12 / 4 − 5 2 + (48/24) − 10 + (12/4) − 5 2 + 2 − 10 + 3 − 5 2 + 2 + (−10) + 3 + (−5) −8
• • • •
Reasons Begin here Group all the multiplications Do all the multiplications Group all the divisions Do all the divisions Convert all the subtractions to negative additions Do the additions from left to right
Group all the divisions Do all the divisions from left to right Convert all the subtractions to negative additions
Do the additions from left to right
When you use these rules correctly. paying careful attention to signs and using parentheses when we need them: 5 × (−4) = −20 −20/(−2) = 10 10 × 8 = 80 80/(−20) = −4 −4/(−4) = 1
The Commutative Law for Multiplication
In basic arithmetic.The Commutative Law for Multiplication
73
Table 5-1. What do you end up with?
Solution
We can break this down step-by-step. each statement is equal to all the statements above it. and finally divide that result by −4. each expression is equal to the one above it. and division without any parentheses to indicate the order in which the operations should be done.
Here’s a challenge!
Start with the integer 5. As you read downward. then multiply that result by 8. As you read down the left-hand column. multiply by −4. the above problem simplifies as shown in Table 5-1. then divide that result by −2. This is a step-by-step simplification of a complicated expression containing addition.

and d. In formal terms. You don’t even get an integer. b. the product is positive. then a /b = c /d When a rule works in both logical directions. The commutative law also works if there are three or more factors. the product is negative. If a /b = c /d. then ad = bc This rule works in the reverse sense.” Now we know that for any four integers a. a /b and c /d. You can rearrange them in any order you want. You write
a /b = c /d You can multiply the dividend on the left-hand side of this equation (here. You’re also assured that neither b nor d is equal to 0. the order is important. But if you divide −3 by 0. and d. you get an undefined quantity! Cross-multiplication Here’s a fact that you will find useful in algebra. If the factors have opposite signs. But if you divide 4 by 20. where b ≠ 0 and d ≠ 0. the product is negative. If you divide 20 by 4. the product is positive. c. you don’t get 5.74 Multiplication and Division
It works when you multiply The fact that you can multiply two integers in either order and get the same result is called the commutative law for multiplication. that’s a) by the divisor on the right-hand side (d ). and get the same result as when you multiply the divisor on the left-hand side (b) by the dividend on the right-hand side (c). b. you get 0. mathematicians use the expression “if and only if ” and abbreviate it as “iff. It’s called the rule of cross-multiplication. where b is not equal to 0 (written b ≠ 0) and d ≠ 0. For any four integers a. If there are an odd number of negative factors. you get 5. A more dramatic example is the division of 0 by any other integer. Suppose you have two ratios of integers. If there are an even number of negative factors. c.
It fails when you divide When you divide an integer by another integer. If you divide 0 by −3. For any two integers a and b
ab = ba If both factors have the same sign (positive or negative). too. If ad = bc. ad = bc iff a /b = c /d
. and you’re told that they’re equal.

We know this because it is an application of one of those two facts we’re supposed to have “homogenized” earlier in this chapter. divide 20 by 4. involves a product of three factors. What’s the verdict? If we have two nonzero integers a and b whose absolute values are the same. then a /b = b /a
The Associative Law for Multiplication
Another important rule that applies to multiplication involves how the factors are grouped when you have three or more of them. we can write If |a| = |b|. Therefore.
Here’s a challenge!
Under what circumstances can we divide an integer a by an integer b. meaning that they’re either identical or are additive inverses of each other. getting 5. we can substitute a for b and get a /a = a /a which simplifies to1 = 1. we can substitute −a for b in the original equation and get a /(−a) = (−a)/a which simplifies to −1 = −1. Finally. getting 100.
Solution
We have two integers a and b. Here is an example: 200/2/5/4 In a case like this.The Associative Law for Multiplication
75
Are you confused?
Suppose you see an expression where you have to divide repeatedly. then a/b = b/a. and you’ll always end up with the same product. Then divide 100 by 5. called the associative law for multiplication.
. −a = b. getting 20. with no parentheses telling you which division to do first. The simplest case. and get the same quotient (or ratio) as when we divide b by a? Assume that a ≠ 0 and b ≠ 0. Here. In symbols. You can lump the factors together any way you want. That’s trivial! Now suppose that a and b are additive inverses. Here. proceed from left to right. that means you should take 200 and divide it by 2. and we are told that a /b = b /a This equation is always true when a and b are the same. In that case.

000/40 = 100. For any three integers a. and finally 10/5 = 2. as in subtraction. It also works if there are more than three numbers in a product.000/40) / (10/5)
. negative. b. as long as there aren’t infinitely many. Consider this:
(16/4)/2 = 4/2 = 2 but 16/(4/2) = 16/2 = 8 For any three integers a. Let’s try this: 4. b. we get 4.
It fails when you divide In division. and c
(ab)c = a(bc) For instance: (3 × 5) × 2 = 15 × 2 = 30 and 3 × (5 × 2) = 3 × 10 = 30 This works whether the numbers are positive.
Are you confused?
Look at another expression where we have to divide more than once. Now let’s think of it this way: (4. the way in which you group the integers or variables is important. it is not necessarily true that (a /b)/c = a /(b /c) The associative law hardly ever works with division. and c. or 0.76 Multiplication and Division
It works when you multiply Here’s how a mathematician would formally state the associative law in its most basic form. and see what happens when we insert parentheses in different places.000/40/10/5 Going straightaway from left to right. then 100/10 = 10.

First.000/4/5 Starting at the left. Consider this infinite product: 1 × (−1) × 1 × (−1) × 1 × (−1) × 1 × (−1) × ··· If we start multiplying from left to right. −1.000. Now let’s try this: 4. −1. . Evidently we can’t!
.. getting 4. we get 4. 1. 1. Now it seems as if the product of the whole thing is equal to 1! What’s going on?
Solution
We derived a contradiction here because we improperly used the commutative and grouping laws. like this: [(−1) × (−1)] × [1 × 1] × [(−1) × (−1)] × [1 × 1] × ··· That gives us 1 × 1 × 1 × 1 × ··· The sequence of products in this case is 1. The final product therefore cannot be defined. without really knowing if we can get away with such tricks. −1. The products switch back and forth between 1 and −1. 1. infinitely many times. we get a sequence of products that come out like this.000 by 5 gives us 200.000 / (40/10) / 5 We do the division in parentheses first. Then dividing 1. Second.” But suppose we try to apply the commutative law to the original expression infinitely many times! We can then rearrange the factors to get this: (−1) × (−1) × 1 × 1 × (−1) × (−1) × 1 × 1 × ··· Now imagine that we try to group the integers in pairs. 1. 1. −1.000/4 = 1.. .. No integer can have two values “at the same time. we tried to apply these rules to a product that is undefined in its most basic form. we acted as if these rules can be used in a single expression infinitely many times..
77
Here’s a challenge!
Here is a riddle that ought to get your brain running at full speed. in order: 1. 1. it simplifies to 100/2 = 50.The Associative Law for Multiplication
In this case. endlessly.

you can use the distributive laws. To do this. and on the associative law for multiplication of three integers. b.
Statements abc a(bc) a(cb) (cb)a cba Q. The right-hand distributive law of multiplication over addition says that (a + b)c = ac + bc
. Table 5-2 is an S/R proof. 4. show that for any three integers a.E.78
Multiplication and Division
Table 5-2. Simply change all the instances of addition to multiplication. and get the same product. The left-hand distributive law of multiplication over addition tells you that this is equal to the product ab plus the product bc. and c arranged so that you must multiply a by the sum of b and c. and c abc = cba
Solution
This works out just like the “challenge” at the end of Chap. b.
Multiplication over addition Suppose you have three integers a. and c are multiplied.
The Distributive Laws
When you come across a sum or difference that is multiplied by a single number or variable. b. Here is a proof that shows how you can reverse the order in which three integers a. As you read down the left-hand column. you will sometimes want to expand it into a sum or difference of products.D. each statement is equal to all the statements above it. Reasons Begin here Group the second two integers Commutative law for the product of b and c Commutative law for the product of a and (cb) Ungroup the first two integers Mission accomplished
Here’s another challenge!
Based on the commutative law for multiplication of two integers. This comes out simpler if you write it down:
a(b + c) = ab + ac Now imagine multiplying the sum of a and b by c.

showing every logical step. b. To see why. each statement is equal to all the statements above it. Reasons Begin here Convert the subtraction to the addition of a negative Left-hand distributive law of multiplication over addition Principle of the sign-changing element Commutative law for multiplication Commutative law for multiplication (again) Principle of sign-changing element (the other way around) Convert the addition of a negative to a subtraction Mission accomplished
. consider it a bonus exercise!
The left-hand law fails with division The left-hand distributive laws do not work for division over addition or for division over subtraction. Warning: Don’t mistake the expression (–1) for the subtraction of 1! The parentheses emphasize that –1 is a factor in a product.D. the process is rather long. for the left-hand distributive law of multiplication over subtraction. all you have to do is produce examples of failure. and c
a(b − c) = ab − ac and (a − b)c = ac − bc It’s not hard to show how these follow from the laws for addition.E. As you read down. If you want to do a rigorous job. That’s easy! Consider this:
24/(4 + 2) = 24/6 = 4 but 24/4 + 24/2 = 6 + 12 = 18 And this: 24/(4 − 2) = 24/2 = 12
Table 5-3. If you want to do a proof for the right-hand distributive law of multiplication over subtraction. Table 5-3 breaks the derivation down into an S/R process. For any three integers a.The Distributive Laws
79
Multiplication over subtraction The distributive laws also work with subtraction.
Statements a(b − c) a[b + (−c)] ab + a (−c) ab + ac (−1) ab + a (−1)c ab + (−1)ac ab + (−ac) ab − ac Q. Derivation of the left-hand distributive law for multiplication over subtraction.

80 Multiplication and Division
but 24/4 − 24/2 = 6 − 12 = −6
The right-hand law works with division If you have a sum or difference as the dividend and the single number as the divisor. narrative proofs are often preferred by mathematicians. then
(a + b)/c = a /c + b /c and (a − b)/c = a /c − b /c
Are you confused?
To help get rid of possible confusion about how the distributive laws operate when integers are negative. Now we can rewrite this as e (c + d ) The left-hand distributive law for multiplication over addition allows us to rewrite this as ec + ed
. First. b. Even though S/R proofs look neat. and if c is any nonzero integer. and c = −4. you can use the distributive law for division over addition or division over subtraction. and d (a + b)(c + d ) = ac + ad + bc + bd
Solution
Let’s do this as a narrative. prove that for any four integers a. We’ll start with the left-hand side of the above equation: (a + b)(c + d ) Let’s think of the sum a + b as a single unit. c. work out the expression where you multiply a times (b + c): −2 × [−3 + (−4)] = −2 × (−7) = 14 Now work out the expression where you add ab and ac: [−2 × (−3)] + [−2 × (−4)] = 6 + 8 = 14
Here’s a challenge!
With the aid of the commutative and distributive laws. and call it e. b = −3. try an example where a = −2. If a and b are any integers.

then divide that result by −25. Evaluate the following expression: 4 + 32 / 8 × (−2) + 20 / 5 / 2 − 8 4. You may (and should) refer to the text as you solve these problems. one step at a time. and d abcd = dcba Here’s a hint: you solved a problem like this for addition in Chap. getting (a + b)c + (a + b)d We can apply the right-hand distributive law twice. Find something more interesting!
. then divide that result by −81. because we just got done “morphing” it using known tactics. The solutions in the appendix may not represent the only way a problem can be figured out. Start with the integer −15. obtaining ac + ad + bc + bd We know this is equal to the expression we began with. and rewrite this as ac + bc + ad + bd Now we employ the commutative law in a generalized way.E. A.
Practice Exercises
This is an open-book quiz. then multiply that result by −9. Don’t hurry! You’ll find worked-out answers in App. multiply by −45. b = 1. and c = 1. Therefore (a + b)(c + d ) = ac + ad + bc + bd Q. c. b. b. 4.Practice Exercises
81
Next. we expand e back into its original form and substitute it into the above expression twice. by all means try it! 1. How does the absolute value change if you start with an integer and multiply by −3 over and over without end? 3. What do you end up with? 6. If you think you can solve a particular problem in a quicker or better way than you see there. Show at least one situation where you can say that a /(b /c) = (a /b)/c where a. Do an S/R proof showing that for any four integers a. 5.D. and finally multiply that result by −5. This proof proceeds in the same way. Do not use the trivial case a = 1. How does the absolute value change if you multiply an integer by −3? 2. and c are integers.

Suppose you have learned the left-hand distributive law for multiplication over addition. the right-hand distributive law for multiplication over addition will always work: (m + n)p = mp + np Use the narrative form. Find something more interesting! 8. Do not use the trivial case a = 1. put together an S/R table showing that for any two integers h and k −(h − k) = k − h
. Show that for any three integers n. and c = 1. you can simply switch the order of the subtraction. but you have never heard about the right-hand version. not an S/R table. and c are integers. b. Construct an S/R table showing that for any two integers d and g −(d + g) = −d − g 10. Prove that when you want to find the negative of the subtraction of one quantity from another.82 Multiplication and Division
7. Show at least one situation where you can say that (ab)/c = a(b /c) where a. b = 1. 9. To do this. and p. m.

You divide the numerator by the denominator.” It means that the fraction can’t be changed into an integer plus or minus a fraction. Otherwise. but in everyday usage. getting the whole-integer part. 6-1. cut distance from 0 in half First. and you’ll still have the same numerical value. In the situation shown by Fig. “It’s 7/3 times as far to Happyville as it is to Bluesdale. such a fraction can seem bizarre. you end up at a point that doesn’t correspond to an integer. the fraction divides out to a plain integer. and the denominator is −2. You can change this so the numerator is −3 with a denominator of 2. the absolute value of the denominator. the dividend is called the numerator. There’s nothing really inappropriate about this type of fraction. “It’s 2-1/3 times as far to Happyville as it is to Bluesdale.84 Fractions Built of Integers
3 2 1
Start here
-1 Finish here Not at integer point -2 -3
Next. or equal to. When the absolute value of the numerator in a fraction is less than the absolute value of the denominator.” Instead they would say. the word “proper” does not imply anything more technically acceptable than “improper. No one ever says anything like. Then you divide the remainder by the denominator to get the fractional part. You can always take the additive inverse of both the numerator and denominator in a fraction. That’s easier to understand. the numerator you start out with is 3. When the absolute value of the numerator in a fraction is larger than.” When the numerator in a fraction is an exact integer multiple of the denominator. you have a proper fraction. and the divisor is called the denominator. an improper fraction can always be changed to an integer plus or minus a fraction. In this context. A proper fraction can also be called a simple fraction.
.
Improper or proper? In a fraction. some people call it an improper fraction. take additive inverse
Figure 6-1 When you divide 3 by −2 and follow the process on
the number line.

” and read “7/3” as “seven to three. If you wanted to include −1 and 1 in the interval above. It contains the points for all possible fractions between. You should also be acquainted with two other symbols. “The Happyville-to-Bluesdale distance ratio is 7:3. If q is a proper fraction. The “larger than or equal to” symbol looks like a “strictly larger than” symbol with a line under it (≥).” then you’re saying the same thing. there are numbers that can’t be expressed as fractions.
Are you confused?
Proper fractions are always larger than −1 but smaller than 1. “It’s seven-thirds times as far to Happyville as it is to Bluesdale. but not quite. The term “ratio” expresses how two quantities are related in terms of their relative size or value. The set of all proper fractions doesn’t account for all the geometric points above −1 and below 1. 9.” Ratios are always expressed in terms of two integers. You’ll learn more about that in Chap. The interval is shown as a shaded line. If you say. Think of the Happyville-Bluesdale situation again. then q > −1 You can also write −1 < q < 1 and q<1
The proper-fraction interval Figure 6-2 shows the “realm of proper fractions” on the number line. These two terms are almost synonymous. You can write.“Messy” Quotients
85
Fraction or ratio? Sometimes a fraction is called a ratio. Within that range.” reading “7:3” as “seven to three. A mathematician would use the “strictly larger than” inequality (>) and the “strictly smaller than” symbol (<) to express this fact. They’re never expressed as a whole integer plus or minus a proper fraction. but not including. “The ratio of the distance to Happyville compared with the distance to Bluesdale is 7/3. you would write
q ≥ −1 Alternatively. you could write −1 ≤ q ≤ 1 and q≤1
. It’s gray (not black) for a subtle reason. one divided by the other. The “smaller than or equal to” symbol looks like a “strictly smaller than” symbol with a line under it (≤). The same thing is true everywhere along the number line. Ratios of integers can’t account for all the points on a true geometric line. The term “fraction” implies that a particular quantity is a part of some other quantity. “The ratio of the distance to Happyville compared with the distance to Bluesdale is seven to three. If you say.” If you want to make clear that you’re talking about a ratio. −1 and 1.” then you’re using 7/3 as a fraction. You would then write something like. but in a different way. you can use a colon instead of a slash to separate the 7 and the 3.

you can express both of these ratios in infinitely many other ways. is t /c = (2a)/(3a) What if you want to specify the actual number of miles per hour that each vehicle can travel? Suppose the truck can go a maximum of 100 miles per hour on a straight. The ratio of the truck’s highest speed to the car’s highest speed. as a ratio of two integers? What is the truck’s top speed to the car’s top speed. is c /t = (3a)/(2a) where a can be any integer except 0.86 Fractions Built of Integers
Figure 6-2 Proper fractions
correspond to points between −1 and 1 on the number line. as a ratio of two integers?
Solution
The term “half again” means “1-1/2 times as great. and the car can
. In general. which you can write as 2/3 or 2:3.” The ratio of the car’s top speed to the truck’s top speed is therefore 1-1/2 to 1. You would write this as 3/2 or 3:2. the ratio of the car’s highest speed (call it c) to the truck’s highest speed (call it t). 3 2 1 0 -1 -2 -3 Proper fractions
Here’s a challenge!
Suppose a car dealer tells you that a certain sports car has a top speed that’s “half again” as fast as the top speed of a certain pickup truck. The ratio of the truck’s top speed to the car’s top speed is 2 to 3. in general. When you want to express a ratio in the reverse sense. Technically. or 3/2 to 1. switch the numerator and the denominator. level road with no wind. What is the car’s top speed compared to the truck’s top speed. But that’s not a ratio of two integers! The correct way to express the ratio is 3 to 2. The open circles at −1 and 1 indicate that they are not included in the interval. expressed as a ratio between two integers.

it means that the numerator and denominator can both be divided by at least one integer.
Negative denominators A fraction with a negative denominator is okay in theory. Conversely. you can multiply both the numerator and the denominator by −1. So you end up with the same number in a form that is easier to comprehend. called a common factor. In these cases. a = 50 in the above equations: c /t = (3 × 50)/(2 × 50) and t /c = (2 × 50)/(3 × 50)
“Reducing” a Fraction or Ratio
Any fraction or ratio can be expressed in countless ways. the ratio of the truck’s top speed to the car’s top speed is 100/150 or 100:150. 6/10 is not in lowest terms. we have the same fraction in a lower form. −1/(−1) is equal to 1. You can probably imagine “negative three-fifths” without much trouble. Finding common factors When a fraction is not in lowest terms. Why bother with such ugly fractions? Both of the fractions or ratios −3/5 and 3/(−5) have the same numerical value. We can divide both the numerator and denominator by 2 and get 3/5:
(6/2)/(10/2) = 3/5 If we start with −6/(−10). but how about “three negative fifths”? That’s tough for almost everybody.
. That will turn the denominator positive while multiplying the value of the entire fraction by −1/(−1). but it’s hard to think about. A fraction or ratio in this form is said to be in lowest terms or lowest form. which is the multiplicative identity element. and the denominator is positive. For example. Of course. and no remainder will be left in either case. Why not use the one that makes more sense? When you see a fraction or ratio with a negative denominator.“Reducing” a Fraction or Ratio
87
go 150 miles per hour under the same conditions. we can divide both the numerator and the denominator by −2 and get 3/5 again: [−6/(−2)]/[−10/(−2)] = 3/5 If we divide both the numerator and the denominator of a fraction by the same integer and get integers in both places. but one form is considered the most “elegant. You can then write the ratio of the car’s top speed to the truck’s top speed as 150/100 or 150:100.” That’s the form in which the absolute values of the numerator and denominator are both as small as possible.

we attach an extra “factor” of −1 to its product of primes.
Are you confused?
When you try to factor the numerator and denominator of a fraction into products of primes. You can check to see that −7/13 = 210/(−390) by dividing 210 by −7 and −390 by 13. we convert both the numerator and the denominator into products of primes. and attach an extra “factor” of −1 to the denominator. 3. you might find that one or the other is already prime or is the negative of a prime. and 5. We start by factoring both the numerator and denominator into products of primes. and do the same thing with the factors in the denominator. The numerator then becomes
210 = 2 × 3 × 5 × 7 and the denominator becomes 390 = −1 × 2 × 3 × 5 × 13 Next. We remove all the common prime factors from both the numerator and denominator. we might not have it in lowest terms. We can always get the lowest form if we are willing to go through a rather tedious process. you simply change both the numerator and the denominator to their additive inverses. This is the lowest form. and we don’t need any intuition to grind it out. If the denominator is negative. −30. That leaves us with a smaller product of primes in the numerator. and a completely different product of primes in the denominator. A mathematician might call this the “brute-force approach. We finish up by multiplying both the numerator and the denominator by −1 to obtain −7/13. Once we have factored both the numerator and the denominator into products of primes. We multiply all the factors in the numerator together. making sure all the prime factors are positive. We do the same thing with the denominator if it is negative. Maybe both are like that! This means the original fraction is in lowest form. If the original numerator is negative. and the denominator has the extra “factor” −1: (2 × 3 × 5 × 7) / (−1 × 2 × 3 × 5 × 13) The common prime factors are 2. in either case. except when the denominator is negative. You’ll get the same integer. we multiply both the numerator and the denominator by −1.88 Fractions Built of Integers
Getting the lowest form Even after we have reduced a fraction to a lower form. If we end up with a negative denominator. we use these products to build a fraction in which both the numerator and the denominator consist of prime factors. Let’s reduce the fraction 210/(−390) to lowest terms according to this set of rules.” It isn’t elegant but it always works. We remove these from both the numerator and the denominator. If the same prime appears in both the numerator and the denominator.
. getting 7/(−1 × 13) That’s 7/(−13). then that prime is a common prime factor. First. we look at those products closely.

we get b·d -----------------------------------------e·e·f This can be written as bd /e 2f. you know what to do!
. for example. and f are all primes. or 0). and e. where a and c are integers (positive. b 2 means bb. The absolute value of the numerator doesn’t have to be smaller than the absolute value of the denominator. but not much.
Multiplying and Dividing Fractions
When you want to multiply two fractions. e.
Solution
This problem is “messy” but not difficult. a. and then below that. They are −1. b (twice). d. first be sure they are both “pure fractions. negative. and b 3 means bbb. write the denominator out in full. and none of the primes in the numerator and denominator are duplicates. We get −1 · a · b · b · b · c · d · e -----------------------------------------−1 · a · b · b · c · e · e · e · f Now it’s easy to see which factors are duplicated in the numerator and the denominator. If either fraction has a negative denominator. scratch a long dashed line underneath it. The parentheses in the denominator are no longer needed because the minus sign is gone. and let’s put small elevated dots (they can represent multiplication. When we remove these factors from the expressions above and below the dashed line.
A fraction times a fraction To multiply a fraction by another fraction. c. and b and d are positive integers. and no two of them are the same. using variables rather than numerical examples. Reduce the following ratio to its lowest form: −ab 3cde /(−ab 2ce 3f ) An exponent of 2 after any factor means that the factor appears twice. c. Let’s write the numerator out in full. Let’s expand the numerator and the denominator out so we don’t have any exponents.Multiplying and Dividing Fractions
89
Here’s a challenge!
Suppose that a.” That means they should both consist of an integer divided by a positive integer. This ratio is in lowest form because both the numerator and the denominator are products of primes. b. remember!) between the factors to make them easy to tell apart. Dividing one fraction by another is a little more complicated. So. Let’s look at how the processes work. but one fraction should be of the form a /b and the other should be of the form c /d. the process is simple. An exponent of 3 after any factor means that the factor appears three times.

The reciprocal of any integer is equal to 1 divided by that integer. and b and d are positive integers. Therefore (a /b)(b /a) = ab /ab = 1 This shows that the reciprocal of a /b is equal to b/a. multiply the numerators and the denominators:
(a /b)(b /a) = ab /ba The commutative law for multiplication can be used to switch around the factors in the denominator in the right-hand side of this equation. is the quantity by which you must multiply the original integer to get 1. To do that. you might as well wait until the entire process is complete before you worry about lowest terms. Then the reciprocal of a /b is equal to b /a. where a and b are integers and neither of them is equal to 0. (a /b)/(c /d ). If you want to divide the quantity a /b by the quantity c /d .
The reciprocal of a fraction The reciprocal of an integer. The reciprocal of every other integer lies somewhere between (but not including) 0 and 1. and −1 is also its own reciprocal. or else somewhere between (but not including) −1 and 0. Suppose a is an integer. If the multiplication is part of a multiple-step calculation process. so you get (a /b)(b /a) = ab /ab Any nonzero quantity divided by itself is equal to 1. It’s easy to see why this is true when you multiply a /b times b /a. It gets that name from the fact that it’s a fraction made of other fractions! You can also think of it as a ratio of ratios. That means 0 has no reciprocal. Suppose you have a fraction or ratio a /b. 1 is its own reciprocal. you can do it like this:
(a /b)/(c /d ) = (a /b)(d /c) = ad /bc The original expression in this equation. You can reduce the product to lowest terms after you’ve multiplied two fractions.90 Fractions Built of Integers
Once you’ve “prepped” the fractions and you want to multiply them together. you can find the reciprocal of the second fraction (the divisor) and then multiply the first fraction (the dividend) by it. and multiply the individual denominators to get the denominator of the product: (a /b)(c /d ) = ac /bd If either of the original fractions is not in lowest terms.
. multiply the individual numerators to get the numerator of the product. You have already been assured that a ≠ 0 and b ≠ 0. but you don’t necessarily have to. is called a compound fraction.
A fraction divided by a fraction When you want to divide a fraction by another fraction. so you know that ab ≠ 0. also called the multiplicative inverse. the product won’t be either. c is a nonzero integer.

convert it into the equivalent form that has a positive denominator. called the common denominator. you can write it as 6/7 and it represents the same number. where m is an integer and p is a positive integer. you can always convert it to the form m /p. contains all such integer ratios that can exist. If you see 6/(−7). d.” If you stumble across the ratio (−6)/(−7). Suppose you have two fractions a /b and c /d in which a and c are integers. the process is straightforward. Let’s look at these problems from a point of view a little closer to algebra. Write an expression for the final quotient.
Solution
Table 6-1 shows how this can be done. e. and then divide that result by e /f. using variables instead of specific numerical examples.Adding and Subtracting Fractions
91
Table 6-1. You want to add them. each statement is equal to all the statements above it. you can rewrite it as −6/7. As you read down the left-hand column. symbolized R.
Adding and Subtracting Fractions
When you want to add or subtract two integers. with the resulting fraction is in lowest terms. in the form of an S/R derivation. and f all be nonzero integers. you should be sure that neither fraction has a negative denominator. Fractions are more involved. c. You’ve probably had plenty of practice adding and subtracting fractions in arithmetic courses. If one of them does. b. where the denominator is nonzero. for example. Suppose you start with a /b and divide it by c /d. If you have some rational number r. and b and d are positive integers. Derivation of a formula for repeated division of fractions. The set of rational numbers.
Here’s a challenge!
Let a.
Statements [(a /b)/(c /d )]/(e /f ) (ad /bc) / (e /f ) (g /h)/(e /f ) gf /he adf /bce Reasons Begin here Apply the formula for division of a /b by c /d Temporarily let ad = g and bc = h. The term rational in this context comes from the word “ratio. and substitute the new names in the previous expression Apply the formula for division of g /h by e /f Substitute ad for g and bc for h in the previous expression
Are you confused?
Any ratio of two integers.
Getting a common denominator When you have two fractions that you want to add or subtract. is known as a rational number. Then you can modify the fractions so they have the same denominator. It’s always possible to get any ratio of this kind into lowest terms. so you write
a /b + c /d
.

ratio. is called a formula. “It depends. As a final step in the addition of two fractions. But you will sometimes come across situations where it’s better to put a fraction. But you don’t always have to. getting ad /bd + cb /db Now apply the commutative law for multiplication to the numerator cb and the denominator db.92 Fractions Built of Integers
Now multiply a/b by d /d.
Are you confused?
When you add two fractions using the above method. but they are in “higher terms. Add these two products together. (Remember. “Does it matter whether or not a fraction is in lowest terms?” The answer is.) The above expression turns into this: (a /b)(d /d ) + (c /d )(b /b) Next. multiply the products of the fractions on each side of the plus sign. and b and d are positive integers. the result might not be in lowest terms. This particular formula is worth “homogenizing” into your brain! If a and c are integers. you should reduce it.
. Then multiply c /d by b /b. then a /b + c /d = (ad + bc)/(bd ) If you’d rather see it in words.”
A fraction plus another fraction Once you’ve found a common denominator for a sum of two fractions. or proportion in a form other than lowest terms. Multiply the denominator of the first fraction by the numerator of the second.” If you want to express a fraction in the simplest or most “elegant” possible way. By now you must wonder. which is also equal to 1. anything multiplied by 1 is equal to itself. which describes a general solution to a math problem. morphing the above expression into ad /bd + bc /bd This produces a sum of two fractions with the common denominator bd. which is equal to 1. then
ad /bd + bc /bd = (ad + bc)/bd An equation such as this. Simply add the numerators. In the above situation. you can reduce the result to lowest terms. the process goes like this: • • • • Multiply the numerator of the first fraction by the denominator of the second. Divide this sum by the product of the denominators. These two fractions have the same numerical values as the original ones. adding the fractions is easy. and put them over the common denominator.

but it conveys the intended meaning better. You could say that 1/10 of the people have green eyes.
Table 6-2. But if you use 100.Adding and Subtracting Fractions
93
Suppose you want to describe how many people in Country X have green eyes. it doesn’t matter if a fraction. These days. Near the end of this proof parentheses and brackets aren’t enough for grouping of an expression. Reasons Begin here Convert subtraction to addition of a negative Principle of the sign-changing element “Divisive identity element”: substitute −1/1 for −1 Multiplication of fractions to right of plus sign Multiplicative identity element: substitite d for 1d Formula for addition of two fractions. per 100. or the ratio of people with green eyes to all the people in Country X is 1:10.
Solution
Table 6-2 shows how the subtraction formula is derived from the addition formula.000 population in Country X. and the machines don’t care about lowest terms. considering (−1)c as a single quantity Commutative law for multiplication Group elements b and c with parentheses Principle of sign-changing element (the other way around) Convert addition of a negative to subtraction Mission accomplished
. nearly everyone uses computers in complicated calculations.000 population. and the computer will output the data in any form you want. As you read down the left-hand column.D. so braces must be used! They look exactly the same as they braces you use to enclose lists of symbols representing set elements. and on the rules you already know.
Statements a /b − c /d a /b + [−(c /d )] a /b + [−1(c /d )] a /b + (−1/1)(c /d ) a /b + (−1)c /1d a /b + (−1)c /d [ad + b(−1)c]/bd [ad + (−1)bc]/bd [ad + (−1)(bc)]/bd {ad + [−(bc)]}/bd (ad − bc)/bd Q. based on the formula for addition of fractions. but the purpose is different.
Here’s a challenge!
Start with the general equation for adding two fractions: a /b + c /d = (ad + bc)/(bd ) Based on this. ratio. That’s not even close to the lowest form. You might discover that one out of every 10 people has green eyes.000 per 100. In pure theory. You can input the numbers as they are. you’ll have to say that the proportion of people with green eyes is 10. each statement is equal to all the statements above it.000 population as a basis. Derivation of a general formula for the subtraction of one fraction from another. prove that a /b − c /d = (ad − bc)/(bd ) as long as b ≠ 0 and d ≠ 0.E. or proportion is in lowest terms or not.

by all means try it! 1. If two fractions are in lowest terms and they are multiplied by each other. 8. b. Imagine that you have a fraction of the form a /b and another of the form c /d. you found a general expression for [(a /b)/(c /d )]/(e /f ) when a. is the quotient always in lowest terms? If so. If you think you can solve a particular problem in a quicker or better way than you see there. If two fractions are in lowest terms and one is divided by the other. and a third of the form e /f. Is the fraction 231/230 in lowest terms? If not. provide examples of both situations. based on your knowledge of the associative law for integers. another of the form c /d. provide an example. and d are integers. where a. d. If not. and f are all nonzero integers. where a and c are integers. The highest wind speed in a “category 4” hurricane is 155 miles per hour. 7. what is the ratio of the absolute temperature of the boiling point to the absolute temperature of the freezing point. b. The highest wind speed in a “category 1” hurricane is 95 miles per hour. Compare this with the formula you got when you solved the “challenge” problem. and then divided the result by e /f. 5. Don’t use any of the trivial cases where all four of the integers have absolute values of 1. c. Imagine that you have fraction of the form a /b. You may (and should) refer to the text as you solve these problems. 10. is the product always in lowest terms? If so. 4. expressed in lowest terms between two integers? 3. 9. reduce it to lowest terms.94 Fractions Built of Integers
Practice Exercises
This is an open-book quiz. On this scale. Show that the associative law works for multiplication of these fractions. e. and e are integers. provide examples of both situations.
. prove it. and b and d are positive integers. If the product is sometimes in lowest terms but not always. 6. Don’t hurry! You’ll find worked-out answers in App. Based on this information. In one of the “challenge” problems. Show that the commutative law works for multiplication of these fractions. and then divide a /b by the result. What is the ratio of these wind speeds. Give at least one example of a situation in which the following equation is true: (a /b)/(c /d ) = (c /d )/(a /b) where a. If not. A. you’ll know what happens when you divide c /d by e /f first. defined in units called kelvins. and b. c. provide an example. Is the fraction −154/165 in lowest terms? If not. expressed in lowest terms between two integers? 2. the coldest possible temperature is 0. pure water at sea level freezes at 273 kelvins and boils at 373 kelvins (to the nearest kelvin). based on your knowledge of the commutative law for integers. c. prove it. d. You divided a /b by c /d. and f are positive integers. If the quotient is sometimes in lowest terms but not always. reduce it to lowest terms. Now find a general expression for (a /b)/[(c /d )/(e /f )] Once you’ve done this. On the absolute temperature scale. The solutions in the appendix may not represent the only way a problem can be figured out.

Consider the decimal numeral 362. Digits for positive powers of 10 are written to the left of the decimal point.000
10 One order of magnitude 10
3
1. Starting at the decimal point and working toward the left:
(2 × 100) + (6 × 101) + (3 × 102) = 2 + 60 + 300 = 362
.
Continues forever! 10
5
100.000
2
100
10
1
10
10 0 10 –1 10 –2 10 –3 Three orders of magnitude 10 –4 10 –5
1
1/10
1/100
1/1. Let’s break it down.). Digits for negative powers of 10 are written to the right of the point.000
1/10. showing values from 100.000 down to 1/100.96 Decimal Fractions
Figure 7-1 This is part of the
positive rational number line.7735.000
10
4
10.000. It looks like an ordinary period (.000
Continues forever!
Start at the decimal point The “cornerstone” on which any decimal numeral is “built” is the decimal point. The powers of 10 range from 5 down to −5.000
1/100.

than −8. Here are a few examples: • • • • 45. b2. called set A and set B.
Here’s a challenge!
Express the scheme for “building” a decimal numeral in general terms. named as follows: A = {a1. in absolute terms. b1 b2 b3 .. we get 362 + 0.
Solution
Imagine two sets of single-digit numerals.888 is four orders of magnitude larger. am} and B = {b1. it is customary to write a single numeral 0 to the left of the decimal point. Suppose A has m elements and B has n elements.. rather than merely providing examples.8888
If one or both quantities is negative. than 565
−88.7735 = 362. If the number is greater than −1 but smaller than 0.000 = 0. than −56 −0.530 0. They mean to say that the absolute value of one quantity is some power of 10 times bigger or smaller than the absolute value of the other quantity. ..98 Decimal Fractions
Starting at the decimal point and working toward the right: (7 × 10−1) + (7 × 10−2) + (3 × 10−3) + (5 × 10−4) = 7/10 + 7/100 + 3/1.000 + 5/10. then a single 0. b3.56 is two orders of magnitude smaller. in absolute terms.300 is one order of magnitude larger than 4. you should be sure to include the phrase “in absolute terms” so there’s no confusion about the meanings of “smaller” or “larger. you should write the minus sign first.565 is three orders of magnitude smaller.. in absolute terms. a3 a2 a1 .. a3..” When you want to portray a decimal number greater than or equal to 0 but smaller than 1..7735 When we add the whole number to the decimal fraction.. a2.. bn The string of numerals to the left of the point represents the sum SA = (am × 10m−1) + ··· + (a3 × 102) + (a2 × 101) + (a1 × 100)
.. bn} Now imagine these single-digit numerals arranged around a decimal point like this: am . and then the decimal point.7735
Are you confused?
You will sometimes hear scientists talk rather loosely about orders of magnitude. .

294/10. For example. You can also write the above examples as • • • • • 0. where there are always two digits to the right of the decimal point.729 represents 729 × 10−3 0.72941 represents 72. and calculus. if there are three digits.7294 represents 7. You’ll come across it when you study higher algebra. if you want to write them. if there are two digits.000 0.72 represents 72 × 10−2 0.7 represents 7 × 10−1 0. it represents 10ths. Mathematicians like to use the nonitalic.72 represents 72/100 0.941 × 10−5
. A numeral of this sort is called a terminating decimal. and so on.000ths. they represent 100ths. If there’s one digit.729 represents 729/1. It’s a good idea to remember this summation symbol. precalculus. uppercase Greek letter sigma (Σ) to represent sums of many numbers.
Tenths. Any further digits.000
When you have a denominator of 10n.72941 represents 72. The most common example is the notation for dollars and cents.000 0. it is the equivalent of multiplying by 10−n.7 represents 7/10 0. where n is a positive integer. the digits to the right of the decimal point always represent a fraction having a denominator that is some power of 10. as far as you want to go.941/100. We can get fancy and write the above expressions like this: ΣA = (am × 10m−1) + ··· + (a3 × 102) + (a2 × 101) + (a1 × 100) and ΣB = (b1 × 10−1) + (b2 × 10−2) + (b3 × 10−3) + ··· + (bn × 10−n) The entire number is therefore the sum Σ = ΣA + ΣB.7294 represents 7. they represent 1.Terminating Decimals
The string of numerals to the right of the point represents the sum SB = (b1 × 10−1) + (b2 × 10−2) + (b3 × 10−3) + ··· + (bn × 10−n)
99
The entire number is represented by the sum S = SA + SB. or whatever In a terminating decimal. are all ciphers.
Terminating Decimals
All of the decimal expressions you’ll see in everyday situations have a finite number of digits to the right of the decimal point.
• • • • • 0. hundredths.294 × 10−4 0.

forever Physicists or engineers see the above numbers differently.103.72941 0. Instead.000ths. the digits to the left represent an integer. but there’s rarely any reason to do that.580. which is in lowest terms because 31 is prime. For example. then adding any more digits to the right of the 1 will only clutter the page with ciphers (zeros). 1.729410000 0. those extra ciphers are important. that would be 7/20. no matter how many digits there are. but often it is not.7864892022 Instead. If you “chop off ” the digits to the right of the point.35.729410 0. you shouldn’t worry about reducing the fractional part to lowest terms unless the nature of the problem demands it. If reduced to lowest terms. the fractional part is 31/100. in the decimal 66.72941. Let’s not worry about that right now.
. If that’s as far as you want to go. The following numerals all represent exactly the same number to a pure mathematician: 0. 100ths. This fraction might be in lowest terms.100 Decimal Fractions
Don’t be fooled here! These examples don’t represent changes in the order of magnitude. However. Those digits to the right of the point are always supposed to represent 10ths. You won’t often see a numeral like this:
00. You might add ciphers to the left-hand end of the digit string without changing the value. To them.103.72941000 0.7294100 0. the number of digits to the left of the decimal point is always finite. because they represent increasing precision or accuracy. When you write or see a decimal expression.
What about lowest terms? When you see a decimal expression that ends after a certain number of digits to the right of the point.000. and so on.7294100000 ↓ and so on.7864892022 In a decimal numeral. the fractional part is 35/100. they all represent numbers that are very close to each other. which is 0. They’re extra significant figures. commas are not customarily inserted to the right of the point.004.
Commas and extra ciphers In any decimal expression. it would be written as 4. in the decimal 66. those digits always express a fraction with a denominator that is some power of 10. The values approach the last number in either list.580.31.

103 .Endless Decimals
101
Are you confused?
Numbers such as 4.580.580. There are 10 digits here.000.022 / 10. But in theory.
.
Solution
Let’s look to the left of the point first.864.000.7864892022 can be hard to read. and also after every third digit to the right of the decimal point.103 + 7. so you can never reach a spot where every digit further to the right is a cipher.580. also called a nonterminating repeating decimal. It can happen in other cases. The calculator program in a personal computer is excellent for this purpose. Those spaces will make the whole thing easier to read. 786 489 202 2 Be careful when you insert spaces into a numeral! In this example.000 The entire number is the sum of these: 4. the digits to the right of the decimal point can continue forever.022 / 10.000
Endless Decimals
Whenever you write out a decimal expression in the “real world. The above numeral would then look like this: 4.000. note that the spaces between these digits don’t correspond to the places you’d put the commas if you were to express them as a fraction. the lonely digit 2 at the end might confuse some people. because it displays a lot of digits. Also. That means the denominator of the fraction should be denoted as 1 with 10 ciphers after it.103 Now let’s look to the right of the point.)
Here’s a challenge!
Break down 4.892.” it’s always a terminating decimal.103.
Endless repeating decimals Your calculator can give you a glimpse of what an endless repeating decimal.580. This always happens when you divide a prime number larger than 5 by any other prime larger than 5.7864892022 into a sum of a single integer and a single fraction. You can insert spaces on either side of the point. producing the fraction 7.864. looks like. (You’ll see this in the “challenge” example below.103. The string of numbers is one big integer: 4. as well. especially when you see them on a calculator display that does not insert the commas.892.580.000.

Even if she has a Ph. The pattern here is too big for the display. But there is always a pattern whether you can see it or not.. Eventually. and divide out this fraction: 138. people knew that the circumference of a circle is slightly more than 3 times its diameter. and then ask her what fraction you put in. 1/7. tell her it’s the quotient of two integers you entered.125 1/9 = 0.333333333333..999.102 Decimal Fractions
A good way to see the difference between a terminating decimal and an endless repeating decimal is to use the “1/x” key on your calculator and start with 2 for x. They tried to define it as a fractional ratio—that is. Once in a while you’ll come across a situation where an integer is divided by another integer and you can’t see the pattern in the decimal expression because the repeating sequence has too many digits. The fractions 1/3. The answer is “Yes. she will not be able to figure it out. For a long time they thought it was 22/7. Take a calculator that can show 10 digits.297.792/999. 1/8 = 0. watching the results: 1/2 = 0. and 1/8 all work out as terminating decimals.25 1/5 = 0.166666666666.
Endless nonrepeating decimals You might wonder whether there are any decimal numbers that go on forever with digits to the right of the decimal point. no matter how large or small the circle happens to be...999. In ancient times. but that don’t produce a repeating sequence.5 1/3 = 0. 1/5.142857142857. mathematicians
.999 Now suppose you show your calculator display to a friend. 1/6. 1/4.. and 1/9 divide out as endless repeating decimals. which goes through a repeating cycle of the six digits 142857. Consider the circumference of a perfect circle divided by its diameter. 4.. The fractions 1/2. Note the uniqueness of 1/7. 1/7 = 0.
Are you confused?
When you divide an integer by another integer. your calculator display might not show enough digits to let you see the pattern of repetition. in math. This value is always the same.111111111111.. and if the two integers are large enough. 5.” Examples are easy to find.004.D.2 1/6 = 0. The presence of an ellipsis (three periods) indicates that the pattern continues forever. Then try it with 3. That’s because any rational number can be expressed as either a terminating decimal or an endless repeating decimal. and so on. 1/4 = 0. as a quotient of two integers—but the best they could do was to come close..

You’ll learn more about them in Chap. Our task is to find a fractional expression for m. divided by 1. Mathematicians call quantities such as π irrational numbers. You’ll never find any repeating pattern of digits in the decimal expansion of π.Endless Decimals
103
were able to prove that this number. are inserted to make the expression clear.. This process is straightforward.) Note that the first addend here is ###/1. Our mission is to show that this decimal numeral represents the fraction ###/999
Solution
Let’s call the “mystery fraction” m. and you shouldn’t have any trouble understanding how it works..000 m = 1. 9. ### ### ### . We can write it down in this form: 0 .000 Now we can multiply each side of this equation by 1.. The second addend happens to be the original mystery number.. like this: m = (0 . m. but it takes several steps. so they’re easy to add. you will fail. We’ve been told that m = 0 . which can be so easily defined in terms of geometry.000/1. The spaces on either side of the decimal point. Even if you spend the rest of your life trying. 000 ### ### ### . We can break this into a sum of two decimal expressions. We get m = (### + m)/1. getting 1. ### ### ### . and after each triplet of pound signs.
Here’s a challenge!
Imagine a decimal expression that has an endlessly repeating triplet of digits.000.000) = ### + m
. Follow along closely.000 = (### + m) (1.000) + (m/1. m = (###/1. one terminating and the other endless.. cannot be defined as a ratio between two integers! If you’ve taken any geometry.000 (### + m)/1.000) The two fractions on the right-hand side of the equals sign have a common denominator. Therefore..000. Many calculators have a key you can punch to get π straightaway. where ### represents the sequence of three digits that repeats. That’s because π is not a rational number. you know that the circumference of a circle divided by its diameter is symbolized by the small Greek letter pi (π). ###) + (0 .000 and then manipulate the right-hand side.

104 Decimal Fractions
Therefore.
Ratio to decimal When you see a ratio of integers. you can convert it to decimal form if you have a calculator that can display enough digits. we divide each side by 999. 1. If you have a calculator with a 10-digit display and you divide 1 by 7. Here’s an example. and the decimal form as a string of digits with a decimal point somewhere. you can always resort to old-fashioned.
Conversions
Every rational number can be expressed in two ways: the ratio form as an integer divided by another integer. If you didn’t know better from having seen the decimal expansion of 1/7 earlier in this chapter. getting m = ###/999 Mission accomplished! Q. But even the best calculators can be overwhelmed if you give them a “bad” enough ratio. you won’t have to perform ratio-to-decimal conversions very often. Terminating decimal to ratio When you see a terminating decimal expression and you want to convert it to a ratio of integers. the calculator program in any good personal computer will usually work.D. If you have a good computer calculator program. you can do it in steps. Try 51/29. obtaining 1. you’re better off. In the extreme. you can always convert it to the other.000m − m = ### + m − m This simplifies to 999 m = ### Finally. If you have a rational number in one form. But that’s the catch! Even a good calculator can fall short in this respect. a computer program to grind out thousands of digits and look for patterns. When you come across a problem where you have to do it. 7601811
.E. Imagine that you are given this decimal numeral and are told to put it into ratio form as a quotient of two integers:
3.000 m = ### + m Subtract m from the expressions on both sides of the equals sign. for example! Fortunately. manual long division.588 . You can also write. you might not be able to deduce it from a 10-digit calculator alone. or find. you will not even see two full repetitions of the pattern.

put 1. This should be easy. In this case you get 35.000.588/1 The third part of the process is a little tricky! Add n ciphers after the 1 in the denominator you just put down.Conversions
105
The extra spaces on either side of the decimal point are there to make it easy to distinguish the digit string to its left from the digit string to its right. you have 7601811.000 If you’d like to check this.000/10. add the fraction you “built” for the whole-number part of the decimal expression to the fraction you “built” for the decimal part.
Endless repeating decimal to ratio The solution to the “challenge” problem in the last section should give you an idea of how to convert any endlessly repeating decimal to a ratio of two integers. That’s a string of seven digits. so the complete ratio is 35.000 + 7.000.880.000. The fractional part is therefore 7.811/10. you multiply the whole-number part of the expression by a certain number and then divide that number by itself. Adding the numerators produces 35.000. In the denominator. so n = 7. Put those digits into the numerator of a fraction. First.000. Put those digits into the numerator of a new fraction.880. Then count the number of digits in the string. When you do this. first split the whole-number part from the decimal part. it’s 10. You can generalize on the number of digits in the repeating pattern.887.601.887. because you have engineered things to get a common denominator! In this case. write a 1 and then n ciphers. In this case.
.000.000 Fourth.811 That’s the numerator of the ratio you want.811/10.811 = 35. In the denominator of the fraction. Suppose that number is n. from one up to as many as you want.000.601. That’s just a fancy (or maybe you’d rather say messy) way of multiplying by 1. The denominator is 10. so that denominator is identical to the denominator you “built” for the decimal part of the expression. In this case. You should get the original decimal expression.601. Then also add n ciphers in the numerator for the whole-number part. divide the ratio out on a calculator that can display at least 11 digits. When you encounter a decimal expression that has a sequence of digits that repeats without end. take the part of the decimal expression to the right of the point. take the part of the decimal expression to the left of the point.000.601. the result is 3.000 Second. Call the whole-number part a.000.

999 × a /999 .” Try reading it over a few times and it should become clearer to you... bk /999 . 999 where the denominator has k digits. and 4 that endlessly repeats..999. so you can easily add to get [(999 .. 6. 999) + (b1 b2 b3 . A specific example.. all 9s. and then put 999 .... The fractional part of the expression is b1 b2 b3 .. with the point on the extreme left: ........ bk b1 b2 b3 . bk where b1 b2 b3 . and it’s easy to “get lost. 999 always stands for a sequence of k digits.. 999 Add this to the fraction you got by converting the decimal part of the original expression.. We can tell right away that this is 8. getting the number in this form: a-b1 b2 b3 . the extra spaces on either side of the decimal point are there only to make it easy to distinguish between the whole-number part of the expression and the decimal part.... That gives you (999 . (Each b with a subscript represents a single digit. bk /999 . getting 999 .. the dash after the a is there only to separate the whole-number part of the expression from the fractional part.106 Decimal Fractions
Then write down the part of the expression to the right of the decimal point in this form. All you have to do is multiply a by the number 999 .
.” can help.604/9. all 9s.. 999 × a) + (b1 b2 b3 . bk /999 . showing the process in “real life action.. 999 × a /999 . Now you can put back the whole-number part... 999 Here. bk b1 b2 b3 . Again. 999 in the denominator... 999..... b1 b2 b3 .... all 9s. 999) You have a common denominator now.. The decimal portion is a sequence of the digits 8.) The extra spaces after the decimal point... are there to make the expression easy to read. 999 Remember that the expression 999 .
Are you confused?
The notation shown above is messy. 0.. and between each digit. bk represents the sequence of k digits that repeats. Let’s convert the following expression to a ratio of integers: 23 .. put the result into the numerator of a fraction.. 860486048604 . It is not a minus sign! Now convert a to a fraction with a denominator consisting of k digits. bk)]/999 ..

581/9.
Here’s a challenge!
There’s a less formal.601. way to do the decimal-to-ratio conversion described in the section “Terminating decimal to ratio” earlier in this chapter.999 = (229. write down a 1 followed by that number of ciphers. and then the sequence of digits 8. How does it work?
Solution
For reference. The denominator should then be 9. The end result might not be in lowest terms. leaving nothing beyond. so we get (23 × 9.Practice Exercises
107
The whole-number portion. Don’t hurry! You’ll find worked-out answers in App. The solutions in the appendix may not represent the only way a problem can be figured out.999 That’s just the whole number 23 expanded into 9.999ths.588 . 6.999.) Then in the denominator of the fraction. The result: 35.604)/9. Count the number of places you moved to the right to get the point to the end of the string of digits. and 4 repeating.977/9.999 If you use a calculator that can display a lot of digits to divide out this fraction. Now we add the decimal part back in.
Practice Exercises
This is an open-book quiz.999 = 238. (In this case. A. Then delete the point.977/9.999. you should get the original expression: 23 followed by a decimal point. and the result put into the numerator of a fraction.999)/9. If you think you can solve a particular problem in a quicker or better way than you see there. You’ll get the whole number 35. but you can reduce it to lowest terms if you want. can be multiplied by 9.887.601. You may (and should) refer to the text as you solve these problems.811 Now make this the numerator of a fraction. but much quicker. 7601811 Move the point to the right until it’s at the end of the string of digits.811/10.999 + 8.977 + 8.000 You can apply this method to any decimal expression you’ll ever see. 23.604/9.887. it’s seven places. here is the original decimal expression again. by all means try it!
. with extra spaces on either side of the point for easy reading: 3.999 = 229. 0. so the entire number becomes 229.000.

7 (b) −8. We use a computer to examine this number to a thousand decimal places. Write the following ratios as decimal expressions. 9. Convert the fraction 1/17 to another fraction whose denominator is a string of 9s. (a) 4. and that the decimal expansion of their ratio (that is. 8. How many orders of magnitude is this? 3.. Write the following decimal expressions as combinations of integers and fractions. with the denominator always positive.000. We are also told that both of these integers are prime numbers. Is there a repeating pattern to the digits in the decimal expansion? 10. 6.892892892 .02 (d) −0.000? Here’s a hint: Express the answer by saying “75. Suppose that somebody tells us there are two integers so large that it would take a person billions of years to write either of them out by hand. Express the numbers from the solutions to Prob. How many orders of magnitude is this? 2. then a billion. then a million. Draw a number line in power-of-10 style that shows the rational numbers from 10 to 100.” where n is a whole number. one of these primes divided by the other) is an endless sequence of digits. How many orders of magnitude larger than 330 is 75.000.. Convert the expression 2. Draw a number line in power-of-10 style that shows the rational numbers from 30 to 300.108 Decimal Fractions
1.000. to a ratio of integers.000 is between n and n + 1 orders of magnitude larger than 330. and we still can’t find a pattern.29 5. 4. 4 as ratios of integers. Can we ever know for sure whether or not a pattern actually exists. Imagine that we come across a gigantic string of digits—miles long if we try to write it out—and we can’t see any pattern. so we can decide whether or not the number is rational? Here’s a hint: This exercise is meant to force your imagination into overdrive!
.35 (c) 0. Reduce the fractional part to lowest terms.000. (a) 44/16 (b) −81/27 (c) 51/13 (d) −45/800 7.

and c ≠ 0 (m /n)0 where m ≠ 0 and n ≠ 0 2 (x − 2x + 1)0 where x ≠ 1 In every expression except the topmost one. and that expression is to be raised to the 0th power. If you’re not sure about the reason for the constraint on x in the last expression. Whenever you find any expression containing variables. These keep the values of the quantities from being equal to 0. Here are some examples of numbers or quantities raised to negative integer powers: 4−2 x− 4 where x ≠ 0 −7 (k + 4) where k ≠ −4 −4 (abc) where a ≠ 0.
a−1 = 1/(a1) = 1/a a−3 = 1/(a 3) a−20 = 1/(a 20) As before. Anything except 0 itself. This is especially important if such an oversight is a step in solving a problem! You would be throwing an undefined quantity into a sensitive process. a can be almost any expression you can imagine. Here are some expressions that are all equal to 1:
40 x0 where x ≠ 0 0 (k + 4) where k ≠ −4 (abc)0 where a ≠ 0. hold on a minute and you’ll see.110 Powers and Roots
The 0th power By convention. Note that the −1st power of any quantity is the same thing as its reciprocal (multiplicative inverse). there are constraints on the variables in the quantities being raised to the 0th power. You’ll also see why 00 is not defined. the expression a n means the reciprocal of the quantity a raised to the power of |n|. and then a complicated expression. You’ll see why any nonzero quantity raised to the 0th power is equal to 1 later in this chapter.
Negative integer powers If a is any number and n is a negative integer. You’ll often see this notation used because it can be a lot less “messy” than writing 1. and c ≠ 0 (m /n)−12 where m ≠ 0 and n ≠ 0 2 (x − 2x + 1)−5 where x ≠ 1
. b ≠ 0. b ≠ 0. be sure you never let that expression attain a value of 0. For example. that is! The quantity 00 is not defined. anything raised to the 0th power is equal to 1. You’ve heard what computer programmers say about putting nonsense into a machine! The same thing happens in mathematics. then a slash.

fourth. third. If you take a negative integer power of 0. as with the 0th power. or c = 0
(k + 4) = 0 (m /n) = 0
But what about the sixth and last expression? It’s not immediately clear that (x 2 − 2x + 1) = 0 when x = 1
You can “plug in” the value 1 for x and see that you get 0 when you add everything up. pay attention to the important message: Beware of taking 0 to the 0th power. you get 21 = 2 22 = 2 × 2 = 4 2 =2×2×2=8
3
2 = 2 × 2 × 2 × 2 = 16
4
↓ and so on. of course! when k = −4 when m = 0 when a = 0.Integer Powers
111
The constraints are imposed. For example. Such equations can be solved in various ways. It turns out that there is one solution to the equation (x 2 − 2x + 1) = 0 That happens to be x = 1.
Here’s a challenge!
What happens if you start with 2 and raise it to successively higher positive integer powers? What happens if you raise 2 to integer powers that get larger and larger negatively?
Solution
If you start with 2 and raise it to positive integer powers. you end up with 1/0. to keep the “powerized” quantities from being equal to 0. Don’t worry about how this type of equation can be solved right now. You’ll learn how to do it later in this book. how the second. 0−7 = 1/(07) = 1/0. Here they are: x=0 (abc) = 0 when x = 0. and fifth quantities above can be equal to 0. without much trouble. and beware of dividing by 0! Don’t let these things happen. thinking it might cause the whole expression to equal 0? Probably not! When you deliberately allow this whole expression to be equal to 0. even accidentally. For the moment. and that’s not defined. But would you have “plugged in” 1 at random. forever
. you get something called a quadratic equation.
Are you confused?
You should be able to see. b = 0.

doubling every time. this is what happens: 2−1 = 1/2 2−2 = 1 / (2 × 2) = 1/4 2 = 1 / (2 × 2 × 2) = 1/8
−3
2 = 1 / (2 × 2 × 2 × 2) = 1/16
−4
↓ and so on. When we take a reciprocal-of-integer power of a quantity.” But if you start with 2 and raise it to integer powers that get larger and larger negatively. but always remaining positive. the result is often called a root. then the pth root of a quantity is something we must multiply by itself p times in order to get that quantity. it “approaches infinity. it is the same thing as taking the pth root of that number. We write this as
a1/p We can surround the exponent with parentheses for clarity. In this particular example. In this case it “approaches 0. If we do that to the above expression. If p is a positive integer. and raise it to the power 1/p where p is a positive integer. becoming half its former size every time you decrease the integer power by 1.
The square root If the general formulas above confuse you.
Integer roots are reciprocal-of-integer powers Suppose we take some number or quantity a. This sequence of values is said to converge. we get a(1/p) In this case. If you have a number and raise it to the power 1/p. let’s find out what goes on when a number is raised to a power that is the reciprocal of an integer. The sequence of values is said to diverge. We know that
52 = 5 × 5 = 25
. forever The value keeps getting smaller and smaller.”
Reciprocal-of-Integer Powers
Now that we’ve seen what happens when a number is raised to an integer power.112 Powers and Roots
The value keeps getting larger without limit. the parentheses are not technically necessary because the whole ratio is written as a superscript anyway. it can help if we look at an example.

or square root. we can say this: If a p = b. so we can say. For any geometric square. to the 3rd power:
43 = 4 × 4 × 4 = 64
Length of edge = s
Interior area = A
A= s2 and 1/2 s=A = A
Length of edge = s
Figure 8-1 The area of a geometric square is equal to
the 2nd power. That’s why the 1/2 power is called the square root. of the length of any edge. We can easily figure out what happens when we raise a number. “The square root of 25 is equal to 5.” By definition then 251/2 = 5 We would say.
.” In general. That’s why the 2nd power is called the square.Reciprocal-of-Integer Powers
113
The second power is often called the square. for any two numbers a and b. The radical consists of a surd symbol (√) with a line extending over the top of the quantity of which the square root is taken. Therefore.
The cube root Now let’s see what happens when p = 3. the length of any edge is equal to the 1/2 power. and for any positive integer p. “5 squared equals 25. the radical notation for square root is shown. the length of any one of the edges is equal to the 1/2 power of the interior area. then b1/p = a The reason the 2nd power is called the square and the 1/2 power is called the square root can be explained in terms of the dimensions and area of a perfect geometric square. in addition to the 1/2 power notation. so 1/p = 1/3. the interior area is equal to the 2nd power of the length of any one of the edges. as shown in Fig. 8-1. say 4. or square. of the area. Looking at it the other way. In the figure.

For any perfect cube. 8-2). We can say. or cube root. “The cube root of 64 equals 4.” Now if we go with the reciprocal power and work backwards. rather than the square root. “4 cubed equals 64. or cube. the length of any edge is equal to the 1/3 power. the volume is equal to the 3rd power of the length of any edge (Fig.114 Powers and Roots
Length of edge = s
Interior volume = V
V = s3 and 1/3 s =V = V
3
Length of edge = s
Length of edge = s
Figure 8-2 The volume of a geometric cube is equal to the
3rd power. of the volume. is indicated by the small numeral 3 in the upper-left part of the radical symbol. we get 641/3 = 4 This can be read as. the length of any edge is equal to the 1/3 power of the volume. That’s because geometric hypercubes having 4 dimensions or more are not commonly named. people write or talk about the numerical powers and roots directly. of the length of any edge. Therefore.
Higher roots When p is a positive integer equal to 4 or more. The fact that the radical refers to the cube root. but you should expect incredulous stares from your listeners if you say “2 tesseracted is 16” or “The tesseract root of 81 is 3.
24 = 16 34 = 81 so 161/4 = 2 so 811/4 = 4
. Going the other way.
The third power is often called the cube. You can check the larger ones on your calculator if you like.” The 3rd power is called the cube and the 1/3 power is called the cube root because of the relationship between the edges and the interior volume of a geometric cube.” Here are some examples of higher powers and roots. The figure also shows the radical notation for the cube root. A 4-dimensional hypercube is technically called a tesseract.

If you use this notation. Therefore.125)1/9 = −5 64 = 1.296?” The answer is “Yes.Reciprocal-of-Integer Powers
115
56 = 15.953.” Both 6 and −6 will work here: 6 × 6 × 6 × 6 = 1.296 so 1. as 11/2 = 1 and −1 Mathematicians get around this problem by saying that whenever “two numbers at once” are the result of a reciprocal power.2961/4 = 6 (−6)4 = 1. a small numeral n is placed in the upper left part of the radical symbol. brackets. so (−1. the positive value is the correct one.296 and (−6) × (−6) × (−6) × (−6) = 1. you must be sure that the radical symbol extends completely over the quantity of which you want to take the root.296 If you multiply any negative number by itself an even number of times.6251/6 = 5 −37 = −2. if you have some number a and its additive inverse −a.187 so (−2. then the pth root of b is ambiguous. If you use the fractional notation.296 so 1. you will get (−a) p = a p every time! If we call (−a)p or a p by some other name such as b.
Are you confused?
Now you will ask.953. and braces should be used to define the quantity of which you want to take the root.187)1/7 = −3 (−5)9 = −1... and then you raise both of those numbers to an even positive integer power p. What? The radical notation can be used for any integer root.125.625 so 15. unless otherwise specified.2961/4 = −6 . “Can 6 and −6 both be valid 4th roots of 1. 161/4 = 2 and −2 811/4 = 3 and −3 15. For the 1/n power. and yet as troubling. That means 161/4 = 2 811/4 = 3
.6251/6 = 5 and −5 It could even mean something as simple. That would mean. parentheses. for example. you’ll get a positive number.

6251/6) = −5 −(11/2) = −1 Sometimes you will actually want to let either the positive or the negative value be used. (−2)4 = (−2) × (−2) × (−2) × (−2) = 16 but −24 = −(24) = −(2 × 2 × 2 × 2) = −16
Negative reciprocal powers We still have not explored what happens when we raise a number to a negative reciprocal-ofinteger power. (−2)3 = (−2) × (−2) × (−2) = −8 and −23 = −(23) = −(2 × 2 × 2) = −8 In contrast to this. If there’s any doubt. be careful if there are no parentheses at all. You can probably figure out the meaning of an expression such as 125−1/3. like this: ±(161/4) = ±2 ±(811/4) = ±3 ±(15. For example. it’s best to place extra parentheses in an expression so everyone knows exactly what it means. Always pay special attention to where the parentheses are placed if you see a negative number raised to a power. In cases of that sort.116 Powers and Roots
15.6251/6) = ±5 ±(11/2) = ±1 Here’s another possible confusion-maker. or
. Also.6251/6 = 5 11/2 = 1 You can indicate that you want to use the negative value by placing minus signs like this: −(161/4) = −2 −(811/4) = −3 −(15. you should throw a plus-or-minus sign (±) into the mix.

and then raise the base a to that power. If you multiply these two quantities. but there are plenty of others. and then take the reciprocal of that. Unlike division by 0 or the 0th root of 0. They have some fascinating properties. which is 5. They’re called imaginary numbers.
Here’s a challenge!
State the rule for negative reciprocal powers in general terms. because a negative divided by a positive always gives us a negative. nothing we know of so far. Mathematicians have defined quantities like this. even roots of negative numbers can be “tamed. This can be any number except 0. Imagine a number and call it a. which is 1/5. and then check out an example. You take the 1/3 power of 125. 21.
Multiply by adding Let’s state the general case first. then we have a−1/p = a− (1/p) = 1/(a1/p)
Multiplying and Dividing with Exponents
When we have a certain base number raised to two different powers. Now imagine the quantities a m and a n.Multiplying and Dividing with Exponents
117
125 to the −1/3 power. where a is the base (the number to be raised to the power) and p is a positive integer. you get the same result as if you add m to n. We will explore them in Chap. Mathematically. where m and n are integers. What can you multiply by itself to get −1? Nothing that we’ve defined yet! What is the 1/4 power of 16? Again. we get two different quantities.” They are commonly used in science and engineering.) If we use the method from the above example where we evaluated 125−1/3.
Solution
The power to which we want to raise the base is −1/p. where p is some positive integer. it goes like this: 125−1/3 = 125−(1/3) = 1/(1251/3) = 1/5
Even roots of negative numbers What happens when you take an even root of a negative number? The simplest example of this sort of problem is the square root of −1. We can write this as an equation:
a ma n = a(m+n)
. But the fact that those quantities have the same base lets us take shortcuts in multiplication and division. (We know that −1/p will be negative. It doesn’t have to be an integer or even a rational number.

Let a = 3. and q = 4. Now we’ll work out an example. and then raise b to that power. Then 105/103 = 100. We can see how it works by trying out an example with specific numbers. and q = 3. Suppose we have b = 10. and n = −4. Then 32 × 3−4 = 9 × 1/81 = 9/81 = 1/9 and 3[2+(−4)] = 3(2−4) = 3−2 = 1/(32) = 1/9
Divide by subtracting Think of a nonzero number. If you divide the first of these quantities by the second. Let b = −2. along with two quantities b p and bq. p = 3. you get the same result as if you subtract q from p.118 Powers and Roots
Let’s call this the addition-of-exponents (AOE) rule.000 = 100 and 10(5−3) = 102 = 100
Are you confused (yet)?
Was that too easy for you? Let’s try a slightly tougher example. Then (−2)3/(−2)4 = −8/16 = −1/2 and (−2)(3−4) = (−2)−1 = 1/(−2) = −1/2
. p = 5. Mathematically:
b p/b q = b(p−q) Let’s call this the subtraction-of-exponents (SOE) rule.000/1. b. m = 2. where p and q are integers.

which has to equal 1 as long as b ≠ 0. Now let’s think of the formula in reverse. so 00 can’t be. So 00 = 0/0 The quantity 0/0 is not defined.
Here’s a challenge!
Using the SOE rule. provide a demonstration of why any nonzero quantity to the 0th power is equal to 1. You might call these facts the generalized addition-of-exponents (GAOE) rule and the generalized subtractionof-exponents (GSOE) rule. we get 00 on the left-hand side of this equation. Then we get 0( p −p) = 0 p/0 p No matter what nonzero value we choose for p. Now we get to the 00 situation. either. We can transpose the left-hand and right-hand sides of the equation to get b( p −q) = b p/b q There’s nothing in the “rule book” that says we can’t have p and q be the same. Let’s do that. Also show why 00 is not defined.
Solution
Look again at the formula that “translates” division of quantities into subtraction of exponents. which must be b raised to the 0th power because p − p is always 0. That formula is b p/b q = b( p −q) where b is the base and p and q are integers. and 0/0 on the right.
. and then raise the result to another power. and call them both p. Let’s violate the “rule book” and let b = 0 in the above equation. In this section we’ll see what happens when you raise a quantity to a power. The right-hand side is b p divided by itself. but for any rational numbers.
Multiple Powers
Numbers can be raised to powers more than once. Then we have b( p−p) = b p/b p The left-hand side of this equation is b raised to the (p − p)th power.Multiple Powers
119
The AOE and SOE rules work not only when the exponents are integers.

(a p)q = a pq Let’s call this the multiplication-of-exponents (MOE) rule.921.0246 = 1.000000000000001 When you have any expression of this sort.976 and [(−0. and then raise the result to the q th power? Mathematically.1)3]5 = (−0. and if p and q are both positive and more than 1. For example. anything to a power) to another power.921. Suppose p and q are integers.
Rational-number powers When you raise an exponentiated quantity (i. and if p and q are both positive and more than 1.504. If the absolute value of a is between 1 and 0.606. The MOE rule still applies.120 Powers and Roots
When exponents multiply Imagine that we have a number a that is not equal to 0. we can let the
.152.504.976 It doesn’t always work out that way. In fact. What happens if we raise a to the pth power.000000000000001 You can evaluate expressions with large exponents quickly by using a calculator with an “x to the yth power” key.606.001)5 = −0.. we get this expression:
(a p)q It is tempting to suppose that the result of this operation will always produce a huge number. That can happen if the absolute value of a is larger than 1.1)3]5 = (−0.1)3×5 = (−0.1)15 = −0. [(−0.e.152. illustrates this: (45)6 = 45×6 = 430 = 1. either or both of the exponents can be negative. Looking at the numerical examples we just saw. however.846. you can get the same result if you take the base a to the power of the product of the exponents pq. the number may be quite close to 0. For example: (45)6 = 1.846. and putting them in this form. That is.

p. you can solve this a lot quicker by noting that 6/3 = 2. so 106/3 = 102 = 100 Usually.0001/3 = 100 If you’re astute. The results often produce numbers that aren’t even rational.Multiple Powers
121
exponents p and q be any rational numbers we want. First. if we encounter an exponent that takes the form r /s.8284 .. But it also tells us something more: when we take a base number to a rational-number. Consider this example: 23/2 = 23×(1/2) = (23)1/2 = 81/2 = 2. It cannot be expressed as a ratio of integers. That gives us the powerful. it’s the same thing as taking the base to an integer power and then taking an integer root of the result. Remember. a rational number is a quotient of two integers! If we reverse the order of the terms in the above three-way equation.
. Remember that a root is a reciprocal power. and q are rational numbers and a ≠ 0. noninteger power. we get a r /s = a r (1/s) = (a r )1/s This is a heavy dose of abstract math! Let’s look at a couple of specific cases where integers are raised to rational-number powers.. rational-number powers aren’t this easy to evaluate. This is an endless nonrepeating decimal. and then use the GMOE rule: (a r )1/s = a r (1/s) = a r /s That’s how we’d evaluate the s th root of a r. far-reaching generalized multiplication-of-exponents (GMOE) rule ! If a.000. we can call this the product of r and 1/s. and is not a rational number. and then take a root of the result. So. You’ll learn more about these types of numbers in the next chapter. then (a p)q = a pq We now have a way to evaluate an expression where we raise a number to a certain power. this: 106/3 = 106×(1/3) = (106)1/3 = 1.

As things turn out. Let’s go: 6(−5/2) = 6−5×(1/2) = (6−5)1/2 = [1/(65)]1/2 = (1/7. It can be tricky because of the minus sign. three ciphers. the expression a−n means you should raise a to the power of |n|. • If a is any nonzero number and m and n are rational numbers.. and there’s no apparent pattern. Then we hit the square root key with the string of digits still in the display. then a m times a n is the same as a raised to the power of (m + n). p. and a long string of digits. First we take the reciprocal of 7. then a m divided by a n is the same as a raised to the power of (m − n). and q are rational numbers and a is nonzero. • If a is any nonzero number and −n is a negative integer. it’s the same as raising a to the power of p /q. • If a. • If a is any nonzero number and p is any nonzero integer. The digits go on without end. getting a decimal point.
. it’s the same as raising a to the power of pq.776. then the pth root of a is the same as raising a to the power of 1/p.122 Powers and Roots
Are you confused?
Let’s review the most important points in this chapter. and q are rational numbers with a and q nonzero. • If a.776)1/2 Now it’s time to use a calculator! Remember that the 1/2 power is the same as the square root. then if you raise a to the pth power and take the result to the qth power. getting 0. • If a is any nonzero number and m and n are rational numbers. this is not a rational number.01134023 . They can be condensed into six statements.
Here’s a challenge!
What do you get if you take the −5/2 power of 6? Mathematically. then if you raise a to the pth power and take
the qth root of the result. p. evaluate this expression and use a calculator to figure out the result to several decimal places: 6(−5/2)
Solution
Let’s apply the GMOE rule to this problem. and we have to be sure we remember the difference between negative powers and reciprocal powers. and then take the reciprocal of the result..

and for any rational numbers p. it is the same as squaring the original number. in narrative form. If you think you can solve a particular problem in a quicker or better way than you see there. ( y + 1)2 5. and r a pa qa r = a ( p +q +r) 7. q. A. How? ( y − 1)2 6. and if r and s are rational numbers. Don’t hurry! You’ll find worked-out answers in App. in the form of an S/R table. The solutions in the appendix may not represent the only way a problem can be figured out. in narrative form. You may (and should) refer to the text as you solve these problems. starting with −1? (b) Do the same thing with a base of −1? (c) Do the same thing with a base of −1/2? 4. 5. Use the GMOE rule to prove. that if x is any number except 0. and for any rational numbers p. Show that if you take the 4th root of any positive number and then square the result. How can we do this? Here’s a hint: Use the results of the final “challenge” in Chap. If we use the same technique as in the previous problem. What happens when a negative number is raised to an even positive integer power? An odd positive integer power? 2. We want to simplify it to a sum of individual terms. that for any number a except 0. then (x r )s = (x s)r 9. Prove. and r [(a p )q ]r = a pqr 8. 10. that for any number a except 0. What happens if we (a) Raise −2 to decreasing negative integer powers. starting with 1? (b) Do the same thing with a base of −1? (c) Do the same thing with a base of −1/2? 3.Practice Exercises
123
Practice Exercises
This is an open-book quiz. Show that if you take the 6th power of any positive number and then take the cube root of the result. What happens if we (a) Raise −2 to increasing integer powers. Use the GAOE rule to prove. by all means try it! 1. it is the same as taking the square root of the original number. Suppose we come across the following expression. q. we can also simplify the following expression.
.

A pattern can never be found that allows us to convert the expression to a ratio of integers.. The decimal number 0. The length of the diagonal of a cube that measures 1 unit on each edge. the expression is an approximation of the actual value. and no rational number is irrational.The Number Hierarchy
1–1/2 1 2
125
1–3/4 1–1/2 2
1–5/8 1–1/2 1–3/4
1–9/16 1–1/2 1–5/8
1–19/32 1–9/16 1–5/8
Figure 9-1 An interval on the rational-number line can be cut in
half over and over. (The last item in the above list has a pattern of sorts.) No matter how many digits we write down to the right of the decimal point. In set notation. No irrational number is rational. The ratio of a circle’s circumference to its diameter. denoted by R. is the union of the set Q of all rationals and the set S of all irrationals:
R=Q∪S
.
• • • •
The length of the diagonal of a square that measures 1 unit on each edge. and you can always find infinitely many numbers in it. S∩Q=∅
Real numbers The set of real numbers.01001000100001000001..
Whenever we try to express an irrational a number in decimal form. but it is not a repeating pattern like the decimal expansion of a rational. This set is disjoint from the set Q of rationals. The set of irrationals can be denoted S. the result is an endless nonrepeating decimal.

or naturals into its “realm. R. until finally you got an interval with P right in the middle. Q. integers. as shown in Fig. and multiplication can be defined over R. 9-1. by introducing the notion of negative values. or multiply a real number by another real number. or take the 0th root of anything and get a real number. we have found out about the set of irrationals. but it’s “standoffish” in the sense that it does not allow any of the rationals. we’ll learn about a set of numbers that’s even larger than the reals. then: x # y ∈R This is a fancy way of saying that whenever you add. Those are the imaginary numbers and complex numbers. It seems reasonable to think that there must be enough rational numbers (there are infinitely many. They result from taking the square roots of negative reals and adding those quantities to other real numbers. Q. We started with the set of naturals. and natural numbers N like this: N⊂Z⊂Q⊂R The operations of addition. This is not generally true of division. take 0 to the 0th power. Now. exponentiation (raising to a power).126 Irrational and Real Numbers
We can envision the reals as corresponding to points on a continuous. Then we built the set of integers. after all) to account for every possible point in this interval. From there. integers Z. in the same way as we can imagine the rationals.
Are you confused?
Take the interval with end points on the number line corresponding to 1 and 2. you can’t take an even-integer root of a negative number and get a real number. by dividing integers by each other. or taking a root.” The Venn diagram of Fig. If # represents any of these operations and x and y are elements of R.
Russian dolls Now we can see the full hierarchy of number types. You can’t divide by 0. subtract. Later on. with a rational number at the middle of each interval. and the set of reals. and R fit inside each other like Russian dolls:
N⊂Z⊂Q⊂R The set S is a proper subset of R. infinitely long line. Z. But there are more points on a real-number line than there are on a rational-number line. But you can’t always do this!
. Also. (Whether or not the real numbers can be paired off one-to-one with the points on a true geometric line is a question that goes far beyond the scope of this book!) The set of real numbers is related to the sets of rational numbers Q. Z. subtraction. straight. The sets N. If that were true. S. N. you always get a real number. then you could take any point P in the interval and slice finer and finer intervals around it. we generated the set of rationals. 9-2 shows how all these sets are related.

when assigned to points on a line. “All right.
irrationals (S). is infinitely more “dense” than the set of rational numbers. How many times more dense is the realnumber line than the rational line? Twice? A dozen times? A hundred times?” The answer might astonish you. but I’ll play along. Here’s a crude analogy. and natural numbers (N). are “dense. as shown in Fig.) The number 21/2 is irrational. The set of real numbers. (That fact comes from basic geometry.”
Are you still confused?
The rational numbers. when depicted as points along a line. rigid
. integers (Z). Both lines resemble straight. known as a unit square. thin. Take a square measuring exactly 1 unit on each edge. rationals (Q). and place it on the number line so one corner is at the point for 0 and the other corner is on the line between the points for 1 and 2.” but not as dense as the points on a line can get.” you say. that line will be full of “holes.
You can define a point between 1 and 2 that doesn’t correspond to any rational number. “This game is strange. The opposite corner of the square falls exactly on the point for the square root of 2. No matter how close together two rational-number points on a line might be. The set of reals is more “dense” than the set of rational numbers. If you build a line using only the points corresponding to rational numbers. 9-3. there is always another rational-number point between them. Suppose you’re exploring the “planet Maths” and you stumble across a rationalnumber line and a real-number line lying in a field. But there are points on a continuous geometric number line that don’t correspond to any rational quantity. infinitely long. because the diagonal of a unit square is precisely 21/2 units long.The Number Hierarchy R = Everything inside the large rectangle
127
N
Q
Z S
S
Q
Z
N
Figure 9-2 The relationship among the sets of real numbers (R).

wires. bullet-by-bullet. the opposite corner is at the point for 21/2. The real-number line contains more “stuff. Some irrational numbers are real. • • • • • Set Q is entirely contained in set R. you find that it’s weightless. all integers are real. The rational-number line is gray. Therefore. The answers. no integers are irrational. Therefore. some irrationals are real.
1 unit
.128 Irrational and Real Numbers Positivenumber line
1 unit
2 This point corresponds to the square root of 2 1
0
Figure 9-3 If you take a unit square and place it diagonally along
the positive-number line with one corner at the point for 0. all irrationals are real.
Solution
We can figure all of these out by looking at Fig. and some Venn diagram reading skill. you discover that it’s heavy. Set Z is entirely contained in set R. Set Z is completely separate from set S. 9-2. Some integers are irrational. Which of the following statements are true. all rationals are real. When you lift it. The real-number line is black. All irrational numbers are real.
Set S is entirely contained in set R. based on our knowledge of the number hierarchy? • • • • • All rational numbers are real. some set theory. Therefore. All integers are real. are as follows. Therefore.”
Here’s a challenge!
Let’s try a little exercise that involves some plain-language logic. Set S has elements in common with set R. When you pick it up. which is irrational. Therefore.

Let’s call those digits a11.More About Irrationals and Reals
129
More About Irrationals and Reals
Let’s explore “infinity” for a few minutes. ℵ0. Then we’ll prove that the square root of 2 is irrational.an1an2an3 . We can keep on listing irrational numbers like this forever. should give you some idea of why this is so. a22. 8. Here is one of them. 3.. and so on. Now imagine that we have listed one irrational number for every possible value of n. 4. (This could not actually be done by any mortal human. and the cardinality of Z is also ℵ0. as we saw in the solution to the final practice exercise in Chap.. This is the same as the number of elements in the set Z of integers. a13. our imaginations let us do infinitely many tasks in a finite amount of time!) Suppose that no two irrational numbers in this list are the same. that is. But in the world of mathematics. To the left of the decimal point. After all. Get into the mood for some serious abstract thinking!
Number lists Mathematicians use the symbol ℵ0 (called aleph-null ) to describe the number of elements in the set N of natural numbers. that set is said to be denumerably infinite (or simply denumerable). 6. as counterintuitive as it may seem. and we haven’t listed any of them twice. So. There are still more irrationals. The first numeral will have an endless string of digits from the set {0.. because it would take forever. 7. we’ll always hit any natural number or integer we care to choose. 3. It is tempting to believe that this list of irrational numbers. 2. can be defined in terms of an “implied list. Suppose we try to list the irrational numbers between 0 and 1 (including 0. If we can make an “implied list” of the elements in an infinite set. a23. the set of rationals. The nth number in our list (corresponding to the nth row in Table 9-1) will be of the form
0. The irrationals and reals can’t be “listed” The elements of S or R cannot be denoted in any type of list. every numeral will have a single 0 and nothing else. Imagine building an irrational number of this form: 0. 5. and by definition it has cardinality ℵ0. the cardinality of Q is the same as the cardinality of N or Z. The second number will have an endless string of digits that we can call a21. there are infinitely many of them. a12. must contain all possible irrationals.
. Table 9-1.b1b2b3 . Cardinality of a set The number of elements in a set is called the cardinality of the set.” Figure 9-4 is an example. We can create “implied lists” of both sets. and be confident that if we go far enough out. But no! Even this infinite list is not complete. The cardinality of N is ℵ0. and so on. We can’t even make an “implied list” of all the nonnegative irrational numbers smaller than 1 in decimal-expansion form. 1. taking the form of a matrix that extends forever to the right and downward from what we see in Table 9-1. and the following explanation. but not 1) by writing down the numerals in their expanded-decimal forms. Even the elements of Q. different from the first string.. 9} to the right of the decimal point.

our new number (with the b’s and subscripts) has at least one digit that doesn’t match. the digit b3 is different from a33 in the third list entry. No matter which of the entries in that list (with the a’s and subscripts) we choose. When we can’t make a list—or even concoct an infinitely long “implied list” in the mathematical cosmos—of all the elements of a set. the digit b2 is different from a22 in the second list entry. you’ll hit the box for any ratio of integers that can exist.
where the digit b1 is different from a11 in the first list entry. starting at the center and going outward as shown.
Are you confused?
The notion of non-denumerability is difficult to grasp. the set is said to be non-denumerable. Don’t feel bad if you don’t fully understand it. This new irrational number cannot be in the list denoted by Table 9-1. and so on forever. Some things that can be defined in
.” Follow the dashed “square spiral” in this two-dimensional list.130 Irrational and Real Numbers
Figure 9-4 All the elements in the set of rational numbers can be
arranged in an “implied list. It’s a little like trying to envision space with more than three dimensions. Eventually.

but that would be a distraction right now. nonrepeating decimal expansions. Let’s call this the integer-squared rule.. think of “infinity” as an expression of the size of a set. It will take some time and effort to understand it. . The cardinality of the set R of real numbers is greater than ℵ0. If we try to create an “implied list” of all the irrational numbers between 0 and 1 as endless.
. bn ≠ ann. and in which problems can be neatly worked out. Cantor was scorned by some of his fellow mathematicians for his theory of transfinite cardinals. Let’s use it now. and therefore that it’s an irrational number. Let’s call this the odd-integer-squared rule. You might wonder if there are any “infinities” larger than the cardinality of the set of reals. Instead. It can help if you stop thinking of “infinity” as something you can count toward. Both of them can be proved. We’ll need two lemmas. . Mathematically. don’t worry. simply defy any attempt at “seeing them in the mind’s eye. for every positive integer subscript n in the table. it’s tempting to try reductio ad absurdum. . the cardinality of N.. Now they are commonly accepted in advanced mathematics. the endless decimal 0. But why not try it? It doesn’t involve anything more complicated or sophisticated than facts of arithmetic you already know.
Here’s a big challenge!
You are now invited to follow along with an “extra-credit” proof.
Solution
Whenever we suspect that a proof might be involved. Each digit in the bottom row (b with subscript) is chosen so that it’s different from the boldface digit above it in the same column (a with subscript). During his lifetime. or Q. If you can’t follow this proof as a whole. Let’s accept them on faith. The mathematician Georg Cantor’s answer to this question was “Yes. • The square of an odd integer is always an odd integer.b1b2b3 . ↓ .More About Irrationals and Reals
131
Table 9-1. • The square of an integer is always an integer. a11 a21 a31 a41 a51 a61 a71 ↓ b1 a12 a22 a32 a42 a52 a62 a72 ↓ b2 a13 a23 a33 a43 a53 a63 a73 ↓ b3 a14 a24 a34 a44 a54 a64 a74 ↓ b4 a15 a25 a35 a45 a55 a65 a75 ↓ b5 a16 a26 a36 a46 a56 a66 a76 ↓ b6 a17 a27 a37 a47 a57 a67 a77 ↓ b7 → → → → → → → ↓→ →
mathematics.
0 0 0 0 0 0 0 ↓ 0 .” It’s good enough to remember that there are more real numbers than rational numbers—a lot more. . Let’s prove that the value of 21/2 cannot be represented as a ratio of integers in lowest terms. we are destined to fail. . Z. . Table 9-2 does the job. Try to understand each step. can’t be in the “a” list. one at a time. infinitely many!” He called these “infinities” transfinite cardinals. even though that list is infinitely long.

132 Irrational and Real Numbers
Table 9-2. but the two expressions represent the same quantity The preceding line contradicts our original assumption about p /q Reductio ad absurdum Mission accomplished
.E. if q were odd. so p2 is an integer p2/2 is an integer p2 is an even integer p is an even integer p /2 is an integer Call p /2 = t p = 2t 2q2 = (2t)2 2q2 = 4t2 q2/2 = t2 We know t is an integer. but q2 is even This follows from the definition of even integer Part of the assumption we made at the beginning of this proof We have proven both of these facts This follows from the statement immediately above this line (p /2)/(q /2) is in lower terms than p /q. if p were odd. p and q. then p2 would be odd. but p2 is even This follows from the definition of even integer This will make things a little simpler Multiply each side of the equation in the previous line by 2 Substitute 2t for p in the equation 2q2 = p2 from earlier in this proof Simplify the right-hand side of the equation in the previous line Divide each side of the equation in the previous line by 4 Integer-squared rule This follows from the previous two lines This follows from the definition of even integer According to the odd-integer-squared rule. so t2 is an integer q2/2 is an integer q2 is an even integer q is an even integer q /2 is an integer The quotient p /q is a ratio of integers in lowest terms p /2 is an integer.
Reasons We begin with this statement and will prove it false Definition of rational number
Assume 2 is rational The quantity 21/2 can be represented as a ratio of two integers.
Statements
1/2
An S/R proof that 21/2 is not a rational number. so q2 is an integer We know p is an integer. and q2 is an integer This follows from the definition of even integer According to the odd-integer-squared rule.D. and q /2 is an integer (p /2)/(q /2) is a ratio of integers The ratio p /q is not given in lowest terms We have produced a logical absurdity 21/2 is not rational Q. then q2 would be odd. in lowest terms 21/2 = p /q (21/2)2 = (p /q)2 2 = p2/q2 2q2 = p2 q2 = p2/2 We know q is an integer.
This is a mathematical statement of the claim made above Square both sides of the equation in the previous line Use arithmetic to manipulate the equation in the previous line Multiply each side of the equation in the previous line by q2 Divide each side of the equation in the previous line by 2 Integer-squared rule Integer-squared rule We know this because q2 = p2/2.

Don’t try to memorize these rules.
x+0=x and 0+x=x
Multiplicative identity For every real number x. and rational-number variables. let’s put together a collection of these properties. integer variables. Just look them over for a little while.How Real Variables Behave
133
How Real Variables Behave
Most of the properties of the natural numbers.
x1 = x and 1x = x
Additive inverse For every real number x. To get ready for the algebra to come. and rationals also apply to the reals. along with a few new or expanded ones. In algebra. But there’s no absolute law about this. integers. real variables are usually given letters from near the end of the alphabet. All the old rules are here. Other letters more often represent natural-number variables (n and m are especially common). Additive identity For every real number x.
x + (−x) = 0 and (−x) + x = 0
. You can always come back to this section for reference if you get stuck later on. especially t and after.
Naming variables You’ll notice that some of the variables here have different letter names than we used before.

the product is equal to 0:
x0 = 0 and 0x = 0
0th power Whenever a nonzero real number x is taken to the 0th power. and never a sum or difference. Therefore (x + y)/z = x /z + y /z and (x − y)/z = x /z − y /z
Zero numerator For all nonzero real numbers x. then the quotient is undefined:
x /0 = (undefined)
Multiplication by zero Whenever a real number x is multiplied by 0. the result is equal to 1:
x0 = 1 when x ≠ 0
. then the quotient is equal to 0:
0/x = 0
Zero denominator For all real numbers x. if 0 is divided by x. if x is divided by 0.How Real Variables Behave
135
These rules also work for subtraction: x (y − z) = xy − xz and (x − y)z = xz − yz Variants of these rules work for division as long as the divisor (or denominator) consists of a single nonzero variable.

Some of the restrictions here.136 Irrational and Real Numbers
Product of signs When two real numbers with plus signs (meaning positive. the following rules apply:
(+)n = (+) (−)n = (−) if n is odd (−)n = (+) if n is even
Reciprocal of reciprocal For all nonzero real numbers. the following rules apply:
(+)(+) = (+) (+)(−) = (−) (−)(+) = (−) (−)(−) = (+)
Quotient of signs When two real numbers with plus or minus signs are divided by each other. the following rules apply:
(+) / (+) = (+) (+) / (−) = (−) (−) / (+) = (−) (−) / (−) = (+)
Power of signs When a real number with a plus sign or a minus sign is raised to a positive integer power. It can help if you work out a few examples using numbers in place of the variables. in which variables are not allowed to equal 0. x. are a little stronger than necessary to keep things straightforward and to be sure we stay safe!
. or larger than 0) or minus signs (meaning negative. or less than 0) are multiplied by each other. n. you don’t have to try to memorize these. As with the facts in the previous section. the reciprocal of the reciprocal is equal to the original number:
1/(1/x) = x when x ≠ 0
More Rules for Real Variables
Here are some more sophisticated rules for expression morphing.

and z. Rational-number powers Suppose that x is a real number. and n ≠ 0. and so on. y.138 Irrational and Real Numbers
Quotient of quotients For all real numbers w. y.
x ( y+z) = x yx z
Difference of powers For all nonzero real numbers x. Then
x m /n = (x m)1/n = (x 1/n)m
Negative powers Let x be a nonzero real number. x. the third root (or cube root) is the same thing as the 1/3 power. and z. Also suppose that m and n are integers.
x yz = (x y)z = (x z)y
. Also suppose that n is a positive integer. x.
w /x + y /z = (wz + xy)/(xz)
Integer roots Suppose that x is a positive real number. y. Then
x−y = (1/x) y = 1/(x y)
Sum of powers For all nonzero real numbers x. The second root (or square root) is the same thing as the 1/2 power. y ≠ 0. Let y be any real number. y. the fourth root is the same thing as the 1/4 power. Then the nth root of x can also be expressed as the 1/n power of x. and z. and z where x ≠ 0. and z where x ≠ 0 and z ≠ 0. y.
x (y − z) = x y/xz
Product of powers For all nonzero real numbers x. (w /x)/(y /z) = (w /x)(z /y) = (w /y)(z /x) = (wz)/(xy) Sum of quotients For all real numbers w. and z ≠ 0.

The value of π.More Rules for Real Variables
139
Quotient of powers For all nonzero real numbers x. For now. Imagine that you stumble across the expression 2π in some mathematical adventure. and z.
x y /z = (x y)1/z = (x 1/z)y
Power of product For all nonzero real numbers x. Let y be any real number.
(x + y)2 = x 2 + 2xy + y 2
Square of difference For all real numbers x and y. is 3.
(xy)z = x zy z
Power of quotient For all nonzero real numbers x. That can include irrationals! You may wonder how can anybody define such a thing as a number raised to the power of pi (π). y.
(x /y)z = x z/y z
Power of reciprocal Let x be a nonzero real number. but you know it can’t be expressed exactly as a ratio of integers. and z. for example. here’s a little glimpse into the mystery. because it
. You’ll understand irrational exponents better when you learn about logarithmic and exponential functions in Part 3.
(x − y)2 = x 2 − 2xy + y 2
Are you confused?
Some of the rules above involve real-number exponents. and z. to five decimal places. y.14159. y. Then
(1/x)y = 1/(x y)
Square of sum For all real numbers x and y.

you do not necessarily get the same result as if you take x to the power of y z.14159 . nonrepeating decimal. If we raise 512 to the 1/2 power.
Solution
All we have to do is find a numerical example where the two expressions don’t agree.. and it corresponds to its own special point on a real-number line. There are plenty! Let x = 2. with x ≠ 0 and y ≠ 0. which is 2. and z = 1/2.. These are both rational numbers. Which of the following quantities can we reasonably suspect is irrational? (a) 163/4 (b) (1/4)1/2 (c) (−27)−1/3 (d) 271/2
. and z are real numbers. an endless. You may (and should) refer to the text as you solve these problems. If we raise x. The solutions in the appendix may not represent the only way a problem can be figured out.
Practice Exercises
This is an open-book quiz. without too much trouble. y = 9. A. You can indeed raise a number. variable.627 . But you can “zoom in” on it. That’s 91/2. Now consider y z. That means 59/19 < π < 60/19 You should now be able to imagine. If you think you can solve a particular problem in a quicker or better way than you see there. 60/19 = 3. and you’ve learned how you can take any number to a rational-number power. we get 22. It’s a real number.... to the power of 3. Then xy = 29 = 512. Divide out the fractions 59/19 and 60/19 with a calculator to obtain their decimal expansions.. Show that if you take x to the yth power and then take the result to the zth power.15789 . we get 8. that 259/19 < 2π < 260/19 The “mystery number” is in that interval somewhere.10526 . you will see that π is between these two rationals: 59/19 = 3.
Here’s a challenge!
Suppose x. and then display π to the same number of places (using a calculator with a π key). or 3. If you expand them to five decimal places. π = 3..140 Irrational and Real Numbers
is irrational. y... by all means try it! 1. Don’t hurry! You’ll find worked-out answers in App. or expression to an irrational power.

The numbers 501/2 and 21/2 are both irrational. other than the trivial case 1 × 831/2. What are these numbers? 6. into a product of a natural number and an irrational number.000. 10. How can we tell? 7. Using the product of sums rule. or any natural-number power of 10? Will the results always be irrational? 3. x. derive a formula for multiplying out the following expression. But the ratio of 501/2 to 21/2 is a natural number. or resolved. What natural number? 8. add the fractions 7/11 and −5/17. or 1. Using the rules we have learned so far. Suppose we have an irrational number and we display the first few digits of its endless nonrepeating decimal expansion. multiply out the product (x + y)(x − y). 9. Using the sum of quotients rule. is the result still irrational? What if we multiply it by 100. and then reduce this sum to lowest terms. If we multiply this number by 10. v. What is the cardinality of (a) The set of even natural numbers? (b) The set of naturals divisible by 10 without a remainder? (c) The set of naturals divisible by 100 without a remainder? (d) The set of naturals divisible by any natural power of 10 without a remainder? 4.Practice Exercises
141
2. Consider this equation in real variables x and y: 36x + 48y = 216 How can we simplify this so it has the minimum possible number of symbols (variables and digits)? 5. y. The nonnegative square root of 83 cannot be resolved into a product of a natural number and an irrational number. and z are real numbers: (u + v + w)(x + y + z)
. w. The nonnegative square root of 18 can be simplified. where u.

Question 1-4
What’s the difference between an “English billion” and a “U. octal.000. and the hexadecimal system works in base 16. Hexadecimal D stands for decimal 13.
Question 1-5
What’s the difference between the decimal. or with systems that used simple counting. Hexadecimal E stands for decimal 14. then the binary system works in base 2.Part One Question 1-3
143
The Hindu-Arabic system uses the numeral 0 to represent a quantity of nothing. Why is this significant?
Answer 1-3
The numeral 0 serves as a placeholder when writing numerals to represent large numbers. and what are their decimal equivalents?
Answer 1-6
Here are the extra digits and their decimal equivalents: • • • • • • Hexadecimal A stands for decimal 10. Eventually. the octal system works in base 8. What are these digits. Hexadecimal F stands for decimal 15.000. possible. Hexadecimal C stands for decimal 12. In the United States. and hexadecimal numeration systems?
Answer 1-5
The decimal system works in base ten.
Question 1-6
The hexadecimal system needs some extra digits besides the usual 0 through 9 to denote numbers.000.000.000. billion”?
Answer 1-4
In England. it made higher mathematics. such as algebra. the term “billion” usually refers to the quantity represented by the numeral 1.
. binary.S. it means the quantity represented by 1.000.000. Hexadecimal B stands for decimal 11. It makes arithmetic easier than it was with the Roman system. The “English billion” is called a “trillion” in the United States. where the numeral 10 represents the number of dots in the following group: •••••••••• If we use decimal numerals to represent quantities.

Chapter 2
Question 2-1
What’s the difference between a set and an element?
Answer 2-1
A set is a collection of elements. 47 is followed by 50. as there usually are in large decimal numerals. 7 is followed by 10. then we write y ∉ X. We can also say that x belongs to X. A set can be an element of another set. what follows 10? What follows 100? What follows 111? What follows 1101?
Answer 1-9
In the binary system. The elements of a set are also called its members.
Question 1-9
In the binary numeration system. If some other number or variable y is not an element of set X. 111 is followed by 1000. 999 is followed by 99A. 10 is followed by 11. what follows 10? What follows FF? What follows 999? What follows FFF?
Answer 1-10
In the hexadecimal system. Note that there are no commas in large binary numerals. and 1101 is followed by 1110. what follows 7? What follows 47? What follows 77? What follows 577?
Answer 1-7
In the octal system.144 Review Questions and Answers Question 1-7
In the octal numeration system. 77 is followed by 100.
Question 1-10
In the hexadecimal numeration system.
. or that x is in X. and 577 is followed by 600. then we write x ∈ X. and FFF is followed by 1. FF is followed by 100. 10 is followed by 11. If we want to call a number or variable x an element of a set X.000. which can be represented by the on/off states of simple switches. 100 is followed by 101. This makes the binary system ideal for use in electronic systems such as computers.
Question 1-8
What is the main advantage of the binary numeration system?
Answer 1-8
It has only two states.

x } C = {y.
P Q
. . What name can we give to the set {x. z. y. one called R and the other called S. z. between the following sets? A = {x. y. y.
Figure 10-1 Illustration for
Questions and Answers 2-4 and 2-5. or even infinitely many times. y. z. set Q. y. However. written R ∩ S. z.}
Answer 2-2
These sets are all the same. z. z. the order of the list isn’t important. if any. and it doesn’t matter. y. z}?
Answer 2-3
The set {x. By definition. x. y. 10-1 represent?
Answer 2-4
This region represents the set of all elements in set P. x } D = {x. the set {x. or 10 times more.
Question 2-4
What does the entire gray shaded region in Fig. They share three elements. called x. z} contains all the elements that belong to both R and S. or z. z. or both. we can list it again. they both contain elements other than x. Once we’ve listed something as an element of a set.Part One Question 2-2
145
What’s the difference. written P ∪ Q. z } B = {z..
Question 2-3
Imagine two sets. z} is the intersection of R and S.. When the elements of a set are listed. and z. That’s the union of sets P and Q. y. y. x } E = {y.

10-1 represent?
Answer 2-5
It portrays the set of elements belonging to both P and Q. all the following sets are identical: X Y X∪Y X∩Y
Question 2-8
Can a set be a subset of itself ? A proper subset of itself ?
Answer 2-8
Any set is a subset of itself. then there are no elements belonging to them both. Their intersection is the set of elements in both X and Y. If X and Y are disjoint. a proper subset of any set S can’t contain all of the elements in S. even itself. But no set is a proper subset of itself. That’s the same as either X or Y by itself. if two sets X and Y are congruent. Their intersection is the set of elements in both X and Y.
Question 2-9
Is there any set that is a subset of every possible set?
Answer 2-9
Yes. Their union is the set of all elements in either X or Y. then they have exactly the same elements. written P ∩ Q.
. The two sets overlap totally. so the intersection is the null set. By definition.146 Review Questions and Answers Question 2-5
What does the hatched region in Fig. In this case. Every element in X is also in Y. That’s also the same as either X or Y by itself. and every element in Y is also in X.
Question 2-7
What’s the union of two congruent sets? The intersection of two congruent sets?
Answer 2-7
By definition. That’s the intersection of the two sets. The null set is a subset of any set we can imagine.
Question 2-6
What’s the union of two disjoint sets X and Y ? What’s the intersection of two disjoint sets X and Y ?
Answer 2-6
The union of two sets X and Y is the set of all elements in either X or Y.

forever
Question 3-2
How can we write the natural numbers 0 through 4 purely in terms of set braces and the null set symbol?
Answer 3-2
We start with 0.. We write this fact as B ⊂ A... 10. That’s because every element in B is also in A. 2} 4 = {0. 1. 8. That’s the null set: 0={}=∅ Once we’ve defined the number 0. How can we do this?
Answer 3-1
We can define the number 0 as the set containing nothing.. 5. n} ↓ and so on. 1. 1. . 11. 7.} B = {7. 1} 3 = {0. 6. 3} ↓ n + 1 = {0. 10.Part One Question 2-10
147
How can we describe the relationship between the following sets? How would we write it in symbolic form? A = {4. 2. which is equal to ∅ by definition. but there are some elements in A that are not in B. 12. 8... like this: 1 = {{ }} = {∅} = {0} After that. . The numbers are built upon each other as sets within sets. 9. B is a proper subset of A. 9. 2..
Chapter 3
Question 3-1
The natural numbers are sometimes defined in terms of sets. like this: 0=∅ 1 = {0} = {∅}
. then we can define the number 1 as the set containing the number 0. we can build the rest of the natural numbers upon each other: 2 = {0. .}
Answer 2-10
In this case.

1. 4. what is a prime number? What’s a composite number? Are there any natural numbers that are neither prime nor composite? Are there any natural numbers that are both prime and composite?
Answer 3-6
A prime number is a natural number larger than 1 (in other words. we symbolize the set of natural numbers as N. 1. 2 or larger) that can only be factored into a product of itself and 1. 3. {∅}. and define it as N = {0. The set N is also known as the set of whole numbers. tell us that the sequence goes on without end. {∅.} This is also called the set of counting numbers. called an ellipsis. {∅.148 Review Questions and Answers
2 = {0. {∅..} The three dots. {∅}} 3 = {0. All the nonprime numbers larger than 1 are composite. We can generate Nodd from Neven by taking every element of Neven and adding 1. {∅}}} 4 = {0. 2. 4. 3.
..
Question 3-6
In the set N. {∅}. {∅}}}}
Question 3-3
How is the set of natural numbers symbolized? What elements does it contain?
Answer 3-3
In this book. We can imagine it as a form of “infinity. 1} = {∅. . 2. 1. 1. the numbers 0 and 1 are neither prime nor composite.. 3. N is defined without including 0: N = {1. {∅. 2} = {∅. 2. what number is represented by the entire set N ?
Answer 3-4
Mathematicians call this an infinite ordinal or transfinite ordinal. and so on. In some texts.”
Question 3-5
How can we generate the set of even natural numbers (call it Neven) from the set N of natural numbers? How can we generate the set of odd natural numbers (call it Nodd) from Neven?
Answer 3-5
We can generate Neven by taking each element of N and multiplying it by 2. and denote it using the lowercase Greek letter omega (ω). According to these definitions.
Question 3-4
According to the set-based definition of the natural numbers 0. {∅}. {∅}}.. 2. No natural number can be both prime and composite. 5. A composite number is a natural number that’s a product of two or more primes. . 3} = {∅.

then it can be factored down further. one-ended list.. Once we’ve found all
. 16. one by one. Any perfect square larger than 1 can be broken down into a product of two or more primes. so we have a whole number. 1. We add 1 to that whole number and call the result s. two-ended list” that can give any reader the basic idea concerning the elements of Z: Z = {. like this: Z = {0. and sometimes it is not. or 10 without a remainder?
Answer 3-9
A natural number is divisible by 2 if its last digit is 0. −2. one-ended list”?
Answer 3-8
Here’s an “implied. No perfect square can be prime. two-ended list”? As an “implied. by 5 if its numeral ends in 0 or 5.Part One Question 3-7
149
In the set N. then we know that the divisor and the quotient are both factors of n. 6. .. 0. −1. or 8. Then we divide the original number n by all the primes less than or equal to s... 1. −1. 5. by 9 if the sum of its numeric digits is a natural-number multiple of 9. 4. so it’s composite.” we start with 0 and then go through the positive and negative integers alternately. We ignore the digits after the decimal point. 25..
Question 3-10
How can we find the prime factors of a large natural number n?
Answer 3-10
We start by finding the square root of n. If it isn’t prime. 3. . and by 10 if its numeral ends in 0. −3. A natural number is divisible by 3 if the sum of its digits is a natural-number multiple of 3. 4. 3. −3. 36. −4. 4. 4..} To create the “implied. Sometimes the quotient is prime. 2. 2.. −2. and 64. 9. If we ever get a whole-number quotient. −4. We keep dividing n by smaller and smaller primes until we get down to 2. 9. The first few perfect squares are 0. 0 and 1 are not prime. 3.}
Question 3-9
How can we quickly and easily tell if a large natural number is divisible by 2. starting with the largest prime and working our way down. By definition. 49.
Question 3-8
How can we write down the set Z of integers as an “implied. 1. 2. what is a perfect square? Can any perfect square be prime?
Answer 3-7
A perfect square is the result of multiplying a natural number by itself.

3 Absolute value is 3 2 1 0 “Number reflector” -1 -2 -3
Absolute value is 3
Figure 10-2 Illustration for Answer 4-1. How can we define a + b on a vertical number line where values increase as we move upward?
Answer 4-2
We start by finding the point on the line that corresponds to a. The absolute value of any natural number is equal to that natural number. If we imagine the integers as points on a straight line with a “number reflector” passing through the point for 0. then the absolute value of any integer is its distance from the “number reflector.
Chapter 4
Question 4-1
What is the meaning of the term absolute value with respect to integers?
Answer 4-1
The absolute value is the extent to which a number differs from 0.150 Review Questions and Answers
the primes smaller than s that divide n without leaving a remainder. Then we move upward along the line for a distance of b units.
Question 4-2
Suppose we have two integers a and b. Some of these prime factors may occur more than once. We end up at the point for a + b. because it is an expression of distance without taking direction into account.
.” Figure 10-2 shows a couple of examples. we have found the prime factors of n. The absolute value of any negative integer can be obtained by removing the minus sign. The absolute value of an integer is always positive or 0.

What’s the result?
Answer 4-5
Let’s work this through in steps. then add −2 to that. the result grows larger. the result grows smaller. then a+b=b+a
. If a and b are integers. For any two integers p and q. paying careful attention to signs and using parentheses when we need them: −6 + (−8) = −6 − 8 = −14 −14 − 12 = −26 −26 − (−5) = −26 + 5 = −21 −21 + (−2) = −21 − 2 = −23 −23 − (−23) = −23 + 23 = 0
Question 4-6
What does the commutative law tell us about the sum of two integers? What does the associative law tell us about the sum of three integers? What do these two laws.
Question 4-4
How do signs work when adding and subtracting positive and negative integers?
Answer 4-4
When we add a positive or subtract a negative. then subtract 12 from that. p + (−q) = p − q and p − (−q) = p + q
Question 4-5
Suppose that we start with −6. Then we move downward along the line for a distance of d units.Part One Question 4-3
151
Suppose we have two integers c and d. and finally subtract −23 from that. When we subtract a positive or add a negative. we can do it in either order and the sum will be the same. We end up at the point for c − d. taken together. allow us to do?
Answer 4-6
The commutative law tells us that when we add two integers. How can we define c − d on a vertical number line where values increase as we move upward?
Answer 4-3
We start by finding the point on the line that corresponds to c. add −8 to it. then subtract −5 from that.

b. Here’s an example showing its failure: (5 − 10) − 15 = −5 − 15 = −20 but 5 − (10 − 15) = 5 − (−5) =5+5 = 10
. then (a + b) + c = a + (b + c) In combination. Let’s look at these two subtractions: 5 − 10 = −5 but 10 − 5 = 5
Question 4-9
Does the associative law work for subtraction done twice?
Answer 4-9
We can’t apply the associative law to subtraction done twice and expect valid results. the commutative and associative laws allow us to arrange and group a sum of integers in any possible way. and the result will always be the same. If a.152 Review Questions and Answers
The associative law says that we can group a sum of three integers in a certain order by twos either way.
Question 4-8
Does the commutative law work for subtraction?
Answer 4-8
We cannot apply the commutative law to subtraction problems and expect valid results. as long as the number of addends is finite.
Question 4-7
Does the commutative law for addition apply to sums of more than two integers? Does the associative law apply for sums of more than three integers?
Answer 4-7
Both of these laws will work for sums having as many addends as we want. and c are integers. and the result will always be the same.

apply the associative law. It’s easy to see this by using a calculator to divide the numerator by the denominator in each of the expressions. and expect valid results in this situation?
Answer 4-10
No! Here’s an example: (5 − 10) + 15 = −5 + 15 = 10 but 5 − (10 + 15) = 5 − 25 = −20
Chapter 5
Question 5-1
When we multiply a positive integer c by another positive integer d. How many times? Give an example.
Question 5-2
What happens if we multiply a positive integer p by a negative integer n? How can we describe that in terms of repeated addition?
Answer 5-2
It works the same way as it does when adding a positive integer. The only difference is that. Can we use parentheses for grouping. For example. as we keep adding the negative integer repeatedly. it’s the equivalent of starting with c and then adding c repeatedly a certain number of times.Part One Question 4-10
153
Suppose a. the sum gets smaller instead of larger. and c are integers and we see the expression a − b + c without any parentheses in it. b. a total of 11 times. a total of 12 times. we get 5 × 12 when we start with 5 and then add 5 over and over.
Answer 5-1
The product cd is equivalent to starting with c and then adding c a total of (d − 1) times.
Question 5-3
In which of the following quotients is there a remainder? (a) 20/10 (e) 105/21
Answer 5-3
(b) 33/11 (f ) 116/29
(c) 51/17 (g) 218/31
(d) 95/19 (h) 301/43
There is a remainder only in case (g).
. Another way to look at this is to imagine that 5 × 12 is what we get when we start with 0 and then add 5 repeatedly.

(c) When we multiply a negative integer by −2 or less.
Question 5-5
Which of the following “rules” is actually false? How can its wording be changed to make it right? (a) When we multiply a positive integer by 2 or more. Convert all the subtractions to negative additions. “The operations of addition. the result becomes positive and the absolute value decreases. or multiply one integer by another. but the operation of division is not. the result becomes positive and the absolute value increases. brackets. the result stays positive and the absolute value increases. (d) When we divide a negative integer by −2 or less.
. Remember that if we multiply a negative by a negative. we get a positive! The correct way to state this rule would be. Do all the divisions.154 Review Questions and Answers Question 5-4
Suppose a mathematician says. the result stays positive and the absolute value decreases. “When we multiply a negative integer by −2 or less. (b) When we divide a positive integer by 2 or more. Suppose we see this: 6 × 8 − 14/2 + 3 What number does this represent?
Answer 5-6
Remember the rules for precedence of operations when we see expressions without parentheses. The steps go in this order: • • • • Do all the multiplications.”
Question 5-6
Sometimes we’ll come across an expression that doesn’t contain parentheses.” What does he/she mean by this?
Answer 5-4
He/She means that we always get an integer if we add. This can be confusing if we don’t know the order in which the operations should be done. the result stays negative and the absolute value increases. subtract.
Answer 5-5
All of the above “rules” are true except (c). and multiplication are closed over the set of integers. But he/she also warns us that we don’t always get an integer if we divide one integer by another. Do all the additions. or braces. subtraction.

. taken together. and finally the last addition? What will the result be then?
Answer 5-7
We should place an opening parenthesis to the left of the 8 and a closing parenthesis to the right of the 14. and the result will always be the same. This gives us 6 × 8 − 14/2 + 3 = 48 − 14/2 + 3 = 48 − 7 + 3 = 48 + (−7) + 3 = 41 + 3 = 44
Question 5-7
How can we change the expression in Question 5-6 to indicate that the subtraction should be done first. and the result will always be the same. then the division. If a. We don’t need to change it to negative addition in this case. Then the subtraction should be changed to negative addition. then (ab)c = a(bc) Taken together. because there’s no risk of ambiguity with the subtraction part alone inside the parentheses. then ab = ba The associative law says that we can group a product of three integers in a certain order by twos either way. allow us to do?
Answer 5-8
The commutative law tells us that when we multiply two integers. and then the ratio. like this: 6 × (8 − 14)/2 + 3 Now we’ve isolated the subtraction problem so it must be done first. We proceed like this: 6 × (8 − 14)/2 + 3 = 6 × (−6)/2 + 3 = −36/2 + 3 = −18 + 3 = −15
Question 5-8
What does the commutative law tell us about the product of two integers? What does the associative law tell us about the product of three integers? What do these laws. and finally the additions should be done.Part One
155
The product should be found first. the commutative and associative laws allow us to arrange and group a product of integers in any possible way. we can do it in either order and the product will be the same. b. and c are integers. then the multiplication. If a and b are integers.

This is −2-1/3. b.156 Review Questions and Answers Question 5-9
Does the commutative law for multiplication apply to products of more than two integers? Does the associative law apply for products of more than three integers?
Answer 5-9
Both of these laws will work for products having as many factors as we want. This is 2-1/2. Imagine three integers a. and c. provided the number of factors is finite. (a) The quotient is 2 with a remainder of 1 (out of 2). (b) The quotient is −2 with a remainder of −1 (out of 3).
Question 5-10
What are the left-hand and right-hand distributive laws of multiplication over addition and subtraction? How do they work?
Answer 5-10
The distributive laws tell us how to work out products when one of the factors is a sum or difference. The left-hand distributive law of multiplication over addition says that a(b + c) = ab + ac The right-hand distributive law of multiplication over addition says that (a + b)c = ac + bc The left-hand distributive law of multiplication over subtraction says that a(b − c) = ab − ac The right-hand distributive law of multiplication over subtraction says that (a − b)c = ac − bc
Chapter 6
Question 6-1
How can we express the following quotients in terms of an integer along with a proper fraction? (a) 5/2 (d) −25/7
Answer 6-1
(b) −7/3 (e) 31/9
(c) 19/2 (f ) 100/(−11)
The problems can be worked out as follows.
.

we remove all the common prime factors from both the numerator and denominator. we can’t apply the commutative or associative laws to quotients or ratios and get valid results.
Question 6-2
Do the commutative and associative laws work for quotients or ratios?
Answer 6-2
Not in general. This is −9-1/11. separating the integers from the fractions. This is −3-4/7. Next. This is 3-4/9. (f ) The quotient is −9 with a remainder of 1 (out of −11). If we end up with a negative denominator.
Question 6-3
What is the “brute force” method of reducing a ratio or fraction to its lowest terms?
Answer 6-3
We begin by factoring both the numerator and denominator into products of primes. That’s the lowest form of 462/561.
. Then we multiply all the factors in the numerator together. (e) The quotient is 3 with a remainder of 4 (out of 9). but we eventually get (2 × 3 × 7 × 11)/(3 × 11 × 17) The common prime factors in the numerator and denominator are 3 and 11. If the original denominator is negative. we attach an extra “factor” of −1.
Question 6-4
How can we use the “brute force” method to reduce 462/561 to lowest terms?
Answer 6-4
First. we do the same thing with it. When we remove these factors. (d) The quotient is −3 with a remainder of −4 (out of 7). we get (2 × 7)/17 Multiplying out the factors in the numerator gives us 14/17.Part One
157
(c) The quotient is 9 with a remainder of 1 (out of 2). This is 9-1/2. This can take a little while. If the original numerator is negative. In most cases. Remember that the short dashes. we factor the numerator and denominator into products of primes. and do the same thing with the factors in the denominator. are not minus signs! The minus signs are the longer dashes. we multiply both the numerator and the denominator by −1. making sure all the prime factors are positive.

That’s because the absolute values of the numerators would be less than the absolute values of the denominators.
. Divide this sum by the product of the denominators. Multiply the denominator of the first fraction by the numerator of the second. In the ratios represented by a.
Question 6-6
How can we quickly add two fractions m /n and p /q. where m and p are integers. 10-3.158 Review Questions and Answers
Figure 10-3 Illustration for
Question and Answer 6-5. b. and n and q are positive integers?
Answer 6-6
We can add these two fractions like this: m /n + p /q = (mq + np)/nq We can also express this in words: • • • • Multiply the numerator of the first fraction by the denominator of the second. That means they’d be improper fractions if written out. and e. the absolute values of the numerators would be larger than the absolute values of the denominators. suppose a through e all represent ratios of integers. 3 2 b 1 c 0 -1 -2 e -3 d a
Question 6-5
On the number line of Fig. Add these two products together. if any. Which of these ratios. would be proper fractions if written out with numerators and denominators?
Answer 6-5
The ratios represented by c and d would be proper fractions if written out.

and multiply the denominators to get the denominator of the product. Then we multiply the numerators to get the numerator of the product. because compound fractions can be awkward and confusing. Subtract the second product from the first. Multiply the denominator of the first fraction by the numerator of the second. p ≠ 0. As a formula: (m /n)(p /q) = mp /nq
Question 6-9
How do we divide a fraction m /n by a fraction p /q. making it q /p.Part One Question 6-7
159
How can we quickly subtract a fraction p /q from a fraction m /n. and n and q are positive integers?
Answer 6-7
The difference can be found this way: m /n − p /q = (mq − np)/nq Stated in words: • • • • Multiply the numerator of the first fraction by the denominator of the second. Divide this difference by the product of the denominators. where m and p are integers.
Question 6-8
How do we multiply a fraction m /n by a fraction p /q. As a formula: (m /n)/(p /q) = mq /np
Question 6-10
What sort of fraction do we have on the left-hand side of the equation in Answer 6-9?
Answer 6-10
This is a ratio of fractions. and multiply the denominators to get the denominator of the product. where m and p are integers. It’s a good idea to simplify expressions like this whenever we can (as the formula above shows). we invert the second fraction.
. and n and q are positive integers?
Answer 6-8
We multiply the numerators to get the numerator of the product. also known as a compound fraction. and n and q are positive integers?
Answer 6-9
First. where m and p are integers.

where n is any positive integer?
Answer 7-1
When we raise a number to the power of 2. The 0th power of 0 is not defined. • When quantities differ by n orders of magnitude... (If we want to add more digits. for example. When we raise a number to the power of 3. This is also called cubing the number. So. After that. In general.160 Review Questions and Answers
Chapter 7
Question 7-1
What does it mean to raise a number to the power of 2? To the power of 3? To the power of n. 3690 = 1 and (−87/16)0 = 1. if b is any number and we raise it to the power of n.g.04
. In the base-10 system: • When quantities differ by one order of magnitude. e. times larger than the absolute value of the other. we have bn = b × b × b × . and then multiply the result by the original number again. we multiply it by itself. • When quantities differ by two orders of magnitude. This is also called squaring the number. compared to another quantity. then the digits will all be ciphers. then the absolute value of one quantity is 102. we multiply it by itself. in terms of absolute value. or 100. to indicate a certain level of accuracy in a physics experiment. then the absolute value of one quantity is 10n times larger than the absolute value of the other. they are no more nonzero digits.
Question 7-4
What’s the difference between a terminating decimal and an endless decimal? What’s the difference between an endless repeating decimal and an endless nonrepeating decimal?
Answer 7-4
A terminating decimal has a finite number of digits to the right of the decimal point. we get 1. then the absolute value of one quantity is 10 times larger than the absolute value of the other. × b (n times)
Question 7-2
What is an order of magnitude in the decimal system?
Answer 7-2
An order of magnitude is a way to express how many times larger or smaller a certain quantity is. but 00 is undefined.
Question 7-3
What is the 0th (or zeroth) power of a nonzero number?
Answer 7-3
When we raise any number except 0 to the power of 0.) Here’s an example: 25-4/100 = 25..

. There are seven digits.000.0056034 into the sum of a whole number and a fraction?
Answer 7-5
We look to the left of the decimal point first.43434343.000 The entire number is the sum of the integer part and the fractional part: 356 + 56.. and the resulting decimal expression approaches (but never quite reaches) the exact value of π.. so we can take them out and add a comma to the digits that remain. We can repeatedly write down the digit pair 43 to the right of the decimal point. But 0. Consider 0. 1/4. the digit sequence 43 repeats forever. or years.14159265.000. which converts to 55/100. 1/10 as decimal expressions?
. but the resulting decimal expression never reaches the precise value of 43/99. This is not a fraction in lowest terms. That means the denominator of the fraction should be 107 or 10. 100ths.55. Now let’s look at an example of an endless nonrepeating decimal: π = 3. We get 0056034/10.000
Question 7-6
When a number is written in decimal form.000.
Question 7-5
How can we convert 356.000. keeping at it for hours.23 converts to 23/100. days..000. and the numerator should be the entire string of digits after the decimal point. is the fractional equivalent (in 10ths. This is in lowest terms.034/10.034/10. but usually not.000ths. getting 56. Now we look to the right of the point. Here..
Question 7-7
How can we write down the fractions 1/2.. but there is no pattern to them. The digits go on forever. or whatever) in lowest terms?
Answer 7-6
Sometimes. The entire string of numbers here is the integer 356. The ciphers at the left in the numerator are useless in the fractional notation. 1.. 1/3. because it can be reduced to 11/20.000 for the fractional part.Part One
161
Here’s an example of an endless repeating decimal: 43/99 = 0. . We can let a computer grind out more and more digits.

8..999
Question 7-9
How can we make a decimal expression an order of magnitude larger or smaller?
Answer 7-9
To make it an order of magnitude larger.0337 To make this number an order of magnitude (or factor of 10) larger.. There are four digits in the sequence. That’s done by moving the decimal point to the right by one place and then repositioning the comma.. all 9s. and 7. 1/8 = 0. we must identify the sequence of digits that repeats.162 Review Questions and Answers Answer 7-7
A calculator can be used for this purpose. so we create a fraction with a denominator having four digits. getting 354. it’s 4. 1/4 = 0.. Here.5 1/3 = 0.587/9..2 1/6 = 0.1
Question 7-8
How can we write the endless repeating decimal 0. 1/7 = 0. we must multiply by 10.458745874587. Then we make the repeating sequence into a four-digit numeral and use it as the numerator of the fraction. move the decimal point to the right by one place. move the decimal point to the left by one place..458745874587.25 1/5 = 0. but it must be able to display a lot of digits! Otherwise. = 4. as a fraction?
Answer 7-8
First.125 1/9 = 0.. Start with 35.166666666666 .680... To make it an order of magnitude smaller. That gives us 0.111111111111 . 1/10 = 0.337
.142857142857 . old-fashioned long division is the best way to find these: 1/2 = 0. Here’s an example.468..333333333333 ... 5.

Part One
163
To make the original number an order of magnitude smaller. which is 1/102 or 0. which is 1. which is 1/103 or 0.546.1. How can we show the meaning of this by examples in the decimal system?
Answer 7-10
Let’s start with the number 10. That can be done by moving the decimal point to the left by one place and then repositioning the comma. which is 0. and so on.401 7−5 = 1/75 = 1/16. we get 10−1. then the −3rd power. and then keeps getting 1/7 as large with each succeeding power.807 ↓ and so on. which is 1/104 or 0. It starts at 1/7. we get 100. This is 101.01 10 divided by itself four times is 10−3. forever
. We end up with 3. When we divide 10 by itself twice.80337
Question 7-10
A negative integer power is a nonzero quantity divided by itself a certain number of times. then the −5th power. then the −4th power.0001 ↓ and so on. It goes on like this for negative integer powers: 10 divided by itself three times is 10−2. When we divide 10 by itself. as far as we want!
Chapter 8
Question 8-1
Suppose we start with 7 and raise it to the −1st power. then the −2nd power. endlessly. we divide by 10.001 10 divided by itself five times is 10−4. Here’s what happens: 7−1 = 1/7 7−2 = 1/72 = 1/49 7−3 = 1/73 = 1/343 7−4 = 1/74 = 1/2. What happens to the result?
Answer 8-1
We get a sequence of fractions that converges toward 0.

Question 8-4
What is meant by the square root of a number? The cube root? The 4th root? The nth root. if we want to find the square root of 16. where n is a positive integer?
Answer 8-4
The square root of a number is a quantity that gives us the original number when squared (multiplied by itself or raised to the 2nd power).164 Review Questions and Answers Question 8-2
When we see a complicated expression raised to negative integer power.
Question 8-5
Even-numbered roots can be ambiguous. and can never attain a value of 0. (d) We can’t let x be equal to y. The 4th root is a quantity that gives us the original number when raised to the 4th power. The nth root is a quantity that gives us the original number when raised to the nth power. That’s forbidden!
Question 8-3
Suppose that we encounter these expressions. we’ll end up dividing by 0. Here’s what we must do to stay safe: (a) We can’t let x be equal to 5. what must we watch out for?
Answer 8-2
We must be sure that the expression is not equal to 0. both 4 and −4 will work. For example. (c) We can’t let either x or y be equal to 0. Otherwise. The cube root is a quantity that gives us the original number when cubed (raised to the 3rd power). (b) We can’t let x be equal to −y. because 42 = 4 × 4 = 16 and (−4)2 = (−4) × (−4) = 16 How can we prevent this sort of confusion?
. What restrictions must we place on x and y in each case? (a) (x − 5)−3 (c) (3xy)−2
Answer 8-3
(b) (x + y)−2 (d) (x − y)−5
We must be sure the expressions inside the parentheses can never equal 0.

How can we express x p/x q as a single power of x ?
Answer 8-8
We find the difference ( p − q).
Question 8-6
What happens when we take a positive odd-integer root of a negative number? A positive even-integer root of a negative number?
Answer 8-6
A positive odd-integer root of a negative number is another negative number. and two integers p and q. How can we express (x p)q as a single power of x ?
Answer 8-9
We multiply p by q. Then we raise x to that power. and two integers p and q. getting x px q = x ( p + q)
Question 8-8
Suppose we have a nonzero number x. Then we raise x to that power. We want to divide x p by x q. Then we insist that if n is even. A positive eveninteger root of a negative number is an imaginary number. We want to multiply x p by x q. and two integers p and q. getting (x p)q = x pq
. the (1/n)th power is positive by default. Then we raise x to that power. we call the nth root of a number the (1/n)th power of that number. We want to raise the quantity x p to the qth power. We can specify that we want to use the negative value instead of the positive one (or along with it. but we must be clear about it. as we’ll do later in this book when we solve quadratic equations). We haven’t worked with imaginary numbers yet.Part One Answer 8-5
165
For any positive integer n. How can we express x px q as a single power of x ?
Answer 8-7
We add the exponents p and q.
Question 8-7
Suppose we have a nonzero number x. getting x p/x q = x ( p − q)
Question 8-9
Suppose we have a nonzero number x.

Another example is π. and then “rolling” it for one complete revolution. infinitely long geometric line and make it into a number line. It’s important that q never be 0! If it is. A good example is the point corresponding to the positive square root of 2.
Question 9-3
How many points on a true geometric number line do not correspond to any rational number? Give two examples. We want to take the q th root of the quantity x p. If those two numbers are p and q. we can always find another rational number between them. like this: (x p)1/q = x p(1/q) = x p /q We divide p by q and then raise x to that power. there are “extra” points on the line that don’t correspond to any rational number.
Chapter 9
Question 9-1
No matter how close together two rational numbers happen to be. an integer p. then the rational number ( p + q)/2 is always greater than p but less than q. 9. the ratio of a circle’s circumference to its diameter. How can we express this as a single power of x ?
Answer 8-10
Let’s remember that the q th root of any quantity is the same as the (1/q)th power. We saw how that point can be located in the text of Chap. and the entire expression will become meaningless. How?
Answer 9-1
We can take the average of the original two numbers. upward along the number line as shown in Fig. This lets us apply the rule from Answer 8-9. Every rational number will then correspond to a unique point on this line. without slipping or sliding. and a nonzero integer q. Does that mean every point on the line will represent a rational number?
Answer 9-2
No! Even though every rational number can be represented on the line. 10-4.
Answer 9-3
There are infinitely many such points. We can locate this point by taking a circle with a diameter of exactly 1 unit.166 Review Questions and Answers Question 8-10
Suppose we have a nonzero number x.
. we’ll get an exponent of p /0.
Question 9-2
Imagine that we take a continuous.

and Q ∩ S = ∅
. • • • • • The set of natural numbers is N. The set of irrational numbers is S. Z ∩ S = ∅. here’s how we symbolize various sets of numbers. and N ⊂ R Z ⊂ Q and Z ⊂ R Q⊂R S⊂R N ∩ S = ∅. The set of rational numbers is Q. N ⊂ Q.
Positive number line This point corresponds to π 3 Circle rolls upward along line without slipping
Finish
2
1
0
Start
1 unit Question 9-4
In this book. The set of integers is Z. The set of real numbers is R.Part One
167
Figure 10-4 Illustration for
Answer 9-3.
Which of these sets are proper subsets of which? Which pairs of sets are disjoint?
Answer 9-4
Here’s how the sets are related: • • • • • N ⊂ Z.

That means the elements of each of these sets can be arranged in an “implied list. (Any integer is equal to itself divided by the integer 1!)
Question 9-6
Which of the sets N. because if we divide a real number by 0. The sets S and R are not denumerable. and subtraction are closed over the set of real numbers. multiplication. S. for any two real numbers x and y. Let’s call that operation “pound” and use the symbol #. The “pound” operation is closed over the set of real numbers if and only if. Division is not. Their elements can’t be arranged in any sort of “implied list. and no imaginary number is real. because 00 is undefined. All integers are real. because that’s the equivalent of taking the number to the power of 1/0. All irrationals are real. Which one? • • • • • • Some real numbers are rational. and Q are denumerable. The 0th root of any real number is undefined.
Question 9-9
What about taking a real root of a real number? Is this operation closed over the set of reals?
Answer 9-9
No. Every integer is rational because it can be expressed as a ratio of integers. Z.
Question 9-8
Which of the common arithmetic operations are closed over the set of reals? Which are not?
Answer 9-8
Addition. the quantity x # y is also a real number. Q. Some integers are irrational.” even though the “list” can’t be written out in full because it’s infinitely long. we get an undefined quantity.” Even an infinitely long “list” can’t capture them all!
Question 9-7
What does it mean for an operation to be closed over the set of real numbers?
Answer 9-7
Imagine that we have an operation between two quantities.168 Review Questions and Answers Question 9-5
One of the following statements is false.
. Any even-integer root of a negative real number produces an imaginary number. and R are denumerable? Which are not?
Answer 9-6
The sets N.
Answer 9-5
The fourth statement is false. Some real numbers are irrational. All integers are rational. Z. Exponentiation (the raising of a real number to a real-number power) is not. No integer is irrational.

The set R has cardinality larger than ℵ0.
. Z. and Q have cardinality of ℵ0 (aleph-null). The sets N.Part One Question 9-10
169
What is meant by the cardinality of a set? What is the cardinality of the set N ? The set Z ? The set Q ? The set R ?
Answer 9-10
The cardinality of a set is the number of elements it contains.

174 Equations and Inequalities
Now suppose we have some variable or quantity. Let’s look at our original numeric example first:
5/2 = 20/8
. Here’s an example with numbers. we get 5/2 − 1/2 = 10/4 − 1/2 = 20/8 − 1/2 Simplifying this gives us the three-way equation 2 = 2 = 2. all parts of an equation and get another valid equation. but we can find it if we want. Let’s call that quantity x. We don’t know its numerical value. We can add x to both sides of any equation we might come across.
Adding or subtracting equations Just as we can add a quantity to. We can subtract the second equation from the first. and further to 4 = 4. or from all parts. Here’s an example: 5/2 = 10/4 = 20/8 If we subtract 1/2 from each part of this. of an equation by the same quantity and get another valid equation. Suppose we see these two equations:
5/2 = 20/8 and 3/2 = 12/8 If we add these two equations to each other by adding their left and right sides individually. we get 5/2 + 3/2 = 20/8 + 12/8 which simplifies to 8/2 = 32/8. we can add or subtract entire equations to or from each other. of an equation and get another valid equation.
Multiplying through by a quantity We can multiply each side. For example. and get 5/2 − 3/2 = 20/8 − 12/8 which simplifies to 2/2 = 8/8. and further to 1 = 1. or subtract it from. or all parts. This is called multiplying through. 5/2 + x = 20/8 + x We can subtract the same quantity from both sides.

Check it out. we get (5/2) × 8 = (20/8) × 8 which simplifies to 40/2 = 160/8.” we say. which is x = −1. We then get x2 + x = 0 Remembering that any quantity squared is equal to that quantity multiplied by itself.
Dividing through by a quantity We can divide any equation through by a nonzero number. and we’ll always get another valid equation. and further to 20 = 20. The danger is worsened by the fact that it can be difficult to tell when such trouble is taking place—until it’s too late. “The problem has been solved. we “plug in” 0 for x in the original and check it out: 02 + 0 + 3 = 3 This simplifies to 3 = 3.
. we divide the equation through by x. As we’ve already learned. we can simplify this to x=0 Confident that we have solved the equation. we’re in trouble. This blinded us to the existence of the other solution. Not so fast! We’ve missed the other solution. That is perfectly legitimate. getting x 2/x + x /x = 0/x Because x /x = 1 and 0 /x = 0 no matter what x might happen to be (or so we think in the excitement of the moment). we must be careful! If that expression can become equal to 0. This oversight occurred because we made the mistake of dividing through by x when one of the two solutions to the original equation is x = 0. when we divide an equation through by some complicated expression that contains one or more variables. As an example of what can happen when we get careless about this. and see what happens. Dividing by any nonzero number is the same thing as multiplying by its reciprocal.Equation Morphing Revisited
175
If we multiply through by 8. Try “plugging in” −1 for x in the original equation. suppose we come across this equation and are told to find the value of x :
x2 + x + 3 = 3 We might start by subtracting 3 from each side.

. You’ve already seen this in action. in general. write 2x ≠ x. It looks like a letter V rotated a quarter-turn counterclockwise.176 Equations and Inequalities
Are you confused?
To avoid mistakes like the one just described. To state or require that x is never equal to y. But sometimes you’ll need to express the fact that quantities differ. It’s an equals sign with a slash through it (≠).
Solution
To see what can go wrong when we square both sides of an equation. consider this: x = −4 This is as simple as an equation can be. which is −4.
Not equal When you want to indicate that two quantities are never equal. The safest approach is to divide through only by nonzero numbers. just as we did when we inadvertently divided by 0. These are statements that involve quantities that are the same. Such statements are called inequalities. or at least the fact that they don’t have to be the same. we’ve generated a false one!
Inequalities
Most of the time in algebra.
Here’s a challenge!
Show that we cannot. instead of missing a legitimate solution. we must never divide an equation through by any variable or expression that might happen to equal 0. It’s clearly not equivalent to the original equation. we get x 2 = 16 This new equation has two solutions. you can use the “not equal to” symbol. and never by variables or expressions containing variables. the “strictly larger than” symbol is used. you’ll work with equations. But this time. or are supposed to be the same. It states its own solution.
Strictly larger When a certain quantity is always larger than (or greater than) some other quantity. write x ≠ y. To state or require that 2x is never equal to x. but you don’t want to specify relative size or the extent to which they’re different. To state or require that x is never equal to 0. If we square both sides. square both sides of an equation and end up with another equation that has the same solution. Here are some examples of its use:
• • • • To state that 3 is not equal to 7/2. We’ve gotten ourselves into trouble. x = −4 and x = 4. write x ≠ 0. write 3 ≠ 7/2.

It looks like a Roman numeral IV rotated a quarter-turn counterclockwise. When you use this symbol. remember that “larger” means “more positive” or “less negative. To state or require that x is strictly larger than y. It looks like a Roman numeral VI rotated a quarter-turn clockwise. When you use this symbol. write −1 ≤ −7/2.Inequalities
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or an arrowhead pointing to the right (>). To state or require that x is larger than or equal to y. write 3 ≥ −7/2.
• • • • To state that −1 is smaller than or equal to 7/2. as is mentioned twice in the above examples? Think for a moment about the meaning of “smaller” with respect to positive and negative numbers. write 2x > x. write x ≥ 0. write 2x < x. To state or require that 2x is smaller than or equal to x.
Larger than or equal When a certain quantity is always larger than or equal to some other quantity. write x ≥ y. To state or require that x is smaller than or equal to y. the “strictly smaller than” symbol is used.
Strictly smaller When a certain quantity is always smaller than (or less than) some other quantity. Then remember what happens when you multiply a negative number by a positive
. the “smaller than or equal” symbol is used. write 2x ≤ x. To state or require that x is larger than or equal to 0. the “larger than or equal” symbol is used. write 2x ≥ x. To state or require that 2x is strictly smaller than x. To state or require that x is smaller than or equal to 0.
Smaller than or equal When a certain quantity is always smaller than or equal to some other quantity. or an arrowhead pointing to the left with a line underneath (≤). or an arrowhead pointing to the left (<). write x ≤ y. To state or require that 2x is strictly larger than x.” • • • • To state that 3 is strictly larger than −7/2. write x < 0. write x > 0. remember that “smaller” means “less positive” or “more negative. It looks like a letter V rotated a quarter-turn clockwise. To state or require that x is strictly smaller than 0. To state or require that 2x is larger than or equal to x. write −1 < 7/2. write x < y. write x ≤ 0. or an arrowhead pointing to the right with a line underneath (≥). write 3 > −7/2. To state or require that x is strictly smaller than y. or smaller than or equal to x.”
• • • • To state that −1 is strictly smaller than 7/2.
• • • • To state that 3 is larger than or equal to −7/2.
Are you confused?
How can a quantity 2x can be strictly smaller than x. To state or require that x is strictly larger than 0. write x > y.

In any logical implication of this kind. then 2x ≤ x Check these facts out with some actual numbers and you’ll see how they work. then 2x is smaller than x. This arrow stands for the term logically implies. It’s not difficult to see that any positive real will work.
Solution
Let’s begin by seeking out all the real numbers that become strictly smaller when we divide them by 10. it’s easy to see that if x is any negative number.” We want the number to get smaller. When any number is negative to begin with. doubling it makes it more negative. the part of the statement to the left of the arrow is called the antecedent. meaning that it becomes larger. not
.178 Equations and Inequalities
number such as 2. then 2x < x and If x ≤ 0. become smaller than the original number. We can say that (x > 0) ⇒ (x /10 < x) If we start with a negative real and then divide it by 10. between the “if ” part of the statement and the “then” part. which in plain English translates to “means it is always true that.
Here’s a challenge!
Write a pair of “if/then” statements that precisely define all the real numbers that. can be abbreviated using a double-shafted arrow pointing to the right.” (It does not mean “causes”!) With the help of this symbol. when divided by 10. An “if/then” statement. the result gets less negative. All the negative reals therefore fail to “qualify. often with a little extra space on either side (⇒). The part of the statement to the right of the arrow is called the consequent. and therefore smaller.
Logical implication Here is a new mathematical symbol. Now you can write If x < 0. the above facts can be shortened to
(x < 0) ⇒ (2x < x) and (x ≤ 0) ⇒ (2x ≤ x) Try reading these statements by saying “logically implies” or “means it is always true that” when you see the arrow. such as those above. Once you remember this.

[(a # b) & (b # c)] ⇒ (a # c)
. Our relation is reflexive if and only if.How Inequalities Behave
179
larger! What about 0? If we divide 0 by 10. Now suppose that we think of some way to compare their values. Such a “comparison tool” is called a relation. the original number must be positive. for all possible values of a and b. b.” Now we can answer the challenge above with a single statement. we use a double-shafted. a. b. But in general mathematics. We can write either
(x /10 < x) ⇔ (x > 0) or (x /10 < x) iff (x > 0)
How Inequalities Behave
Imagine three variables. Let’s define three properties that may (or may not) apply to this relation. for all possible values of a. The antecedent can also be the consequent. We want the number to get strictly smaller! Now. (a # b) ⇒ (b # a) Our relation is transitive if and only if.
Logical equivalence When a logical implication works in both directions. addition to the above statement. we end up with 0 again. double-headed arrow. often with extra space on either side (⇔). In algebra. such as sets or logical statements. To symbolize logical equivalence. they often represent other things. and c. We can also write the cryptic word “iff.
a#a Our relation is symmetric if and only if. the variables represent numbers. This means that the left-hand part of the statement is true if and only if the right-hand part is true. and c. and vice-versa. we have logical equivalence. read as “pound”). so 0 does not “qualify” either. we can claim its reverse: (x /10 < x) ⇒ (x > 0) This means that if we divide a real number by 10 and get a strictly smaller number.
Three properties of relations Suppose we symbolize a newly dreamed-up relation by a pound sign (#. for all possible values of a.

all variables would be different from themselves! The symmetric property. then a > c. 8 in Chap. 11-1. and brackets in logical statements. the various forms of inequality are not equivalence relations. Behavior of the ≠ relation “Not equal” fails the test when it comes to the reflexive property. then # is reflexive. and transitive. however.”
Figure 11-1 The “strictly larger”
relation is transitive. and c: If a is strictly larger than b. if a > b and b > c. We can now write
(∀ a. but it is not true that 1 ≠ 1!
Behavior of the > relation The “strictly larger” relation is neither reflexive nor symmetric. In contrast. symmetric. then a is strictly larger than c. then a > c. just as we use them in ordinary mathematical expressions and equations. we can never say that b >a. then # is called an equivalence relation. Mathematicians symbolize the words “for all” by an upside-down capital letter A (∀).180 Equations and Inequalities
We can use parentheses. Suppose we let a be equal to 1.” If we have a relation # that is reflexive. and transitive.
>
a
>
b
>
c
. Therefore. “For all a.
Behavior of the = relation You may be thinking. so it is an equivalence relation. We can read the above string of symbols out loud as. b. b. c) : [(a > b) & (b > c)] ⇒ (a > c) The colon separates the quantifier from the main substance of the statement. if “not equal” were reflexive. and b is strictly larger than c. The transitive property does work here. and c be equal to 1. we can write this as
(a ≠ b) ⇒ (b ≠ a) “Not equal” can’t pass the transitive property test. The converse of this is also true: If # is an equivalence relation. symmetric. b. we saw that the equality relation (=) is reflexive. If a > b. Then we have 1 ≠ 2 and 2 ≠ 1. and transitive. symmetric. no matter what a and b are. There is no number a such that a > a. 6. and give it the fancy name universal quantifier. If a is not equal to b. “I’ve seen some of these properties before!” In the solution to Prob. Logically. For all numbers a. “strictly larger” is not an equivalence relation. If a > b and b > c. This principle is illustrated in Fig. does work for the “not equal” relation. and c. then b is not equal to a. This is trivial. b be equal to 2. braces. The ampersand (&) stands for “and.

For all numbers a. Because of this. If a ≤ b. b. The “strictly smaller” relation is not an equivalence relation. if a ≥ b and b ≥ c. Therefore. The “strictly smaller” relation is transitive. c) : [(a ≤ b) & (b ≤ c)] ⇒ (a ≤ c)
. simply because it equals itself. 11-2. There is no number a such that a < a. c) : [(a ≥ b) & (b ≥ c)] ⇒ (a ≥ c)
Behavior of the Ä relation The case for the “smaller than or equal” relation is similar to the case for the “larger than or equal” relation. and c. the “strictly smaller” relation cannot qualify as an equivalence relation. We can write
(∀ a. b. We can write these two facts in formal terms as
(∀ a) : a ≥ a and (∀ a. then a < c. then b ≤ a if a = b. but it never works if a is larger than b. but never if a < b. the “larger than or equal” relation is not an equivalence relation. If a < b. however.How Inequalities Behave
181
Figure 11-2 The “strictly smaller”
relation is transitive. c) : [(a < b) & (b < c)] ⇒ (a < c)
Behavior of the ê relation The “larger than or equal” relation is not symmetric. b. just as does the “strictly larger” relation. if a < b and b < c. and that’s good enough! For all numbers a. then b ≥ a if a and b happen to be the same. then a < c. The reflexive and transitive properties hold true for the “larger than or equal” relation. and c. If a < b and b < c. We can logically state these fact as
(∀ a) : a ≤ a and (∀ a. A number a is always larger than or equal to itself.
>
a
>
b
>
c
Behavior of the < relation The “strictly smaller” relation fails the reflexive and symmetric tests. do hold true here. This principle is illustrated in Fig. we can never say that b < a. we can restate the case here by turning all the inequality symbols around. b. b. regardless of the values of a and b. then a ≥ c. If a ≥ b. In fact. The reflexive and transitive properties.

you can read complicated logical statements out loud or write them down in words. let’s make up three statements and give them logical names: • I’m thinking of a natural number = Tn.182 Equations and Inequalities
Are you confused?
Once you know what all these symbols mean. To see why. • I’m thinking of a rational number = Tq. then a is smaller than or equal to c. If I’m thinking of a natural number. we know that Tn ⇒ Tq and Tq ⇒ Tr These statements translate to the following mathematical verses. and reals (N.
Here’s a challenge!
Is logical implication an equivalence relation?
Solution
No. and R) are related. You can even break them up and turn them into “mathematical verse. the last logical sentence in the previous paragraph can be written like this. b. If I’m thinking of a rational number. and c: If a is smaller than or equal to b.” For example. Q. then I’m thinking of a real number. For all a. and b is smaller than or equal to c. • I’m thinking of a real number = Tr.
. rationals. From our knowledge of how the sets of naturals. Now let’s check out the symmetric property. then I’m thinking of a rational number.

We can divide both sides of a “not equal” statement by the same nonzero quantity and get another true statement. Add the same quantity to both sides.) If I’m thinking of a real number.Inequality Morphing
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We can’t reverse either of these implications and still end up with a valid statement. We can add or subtract the same quantity from each side of a “not equal” statement. and a few don’t work at all. then we end up with 0 ≠ 0. then for any number c
a+c≠b+c and a−c≠b−c We cannot.
Inequality Morphing
Some of the familiar equation-morphing rules also work for inequalities.) You can try out some examples if you want to improve your understanding of how they work. If the quantity is 0.) Logical implication is not an equivalence relation. add two “not equal” statements and get another “not equal” statement. we get another valid equation. If I’m thinking of a rational number. (We won’t get into formal proofs of these facts. we can express that fact in either order. but when we add them (left-to-left and right-to-right). Divide both sides by the same nonzero quantity. then b ≠ a. Now let’s see how well these rules work for inequalities. (Suppose it’s the positive square root of 2. it is always true that if a ≠ b. Consider 3 ≠ 4 and 8 ≠ 7.
Whenever we do one or more of these things to an equation. If two quantities are different to start out with. • • • • • • Reverse the order. Here are the things we can do with two-part equations. because it’s not symmetric. If a ≠ b. Add one equation to another. then they’ll still be different if we add or subtract the same quantity from both.
Manipulating ñ statements If two quantities a and b are different. (Suppose it’s 1/2. If the quantity by which we divide through is 0. we get 3 + 8 ≠ 4 + 7. In general. Multiply both sides by the same quantity. These are both true statements. we get undefined results on both sides of the inequality symbol. which is false. But they are equal! We can multiply both sides of a “not equal” statement by the same nonzero quantity and get another true statement. Subtract the same quantity from both sides. in general. summarized for reference.
. I’m not necessarily thinking of a rational number. I’m not necessarily thinking of a natural number. but others must be modified.

then for any number c
a+c>b+c and a−c>b−c We can always add two “strictly larger than” statements (left-to-left and right-to-right) and get another “strictly larger than” statement. Can we add the same quantity to both sides? Yes. We can add or subtract the same quantity from each side of a “strictly larger than” statement. if we have a > b and c > d. we can reverse the order if we also reverse the sense of the inequality. the sense of the inequality is reversed. b.
Manipulating > statements If some quantity a is strictly larger than another quantity b. which is false. Can we add one statement to another? Not in general. • • • • • • Can we reverse the order? Yes. If a > b. then for any positive number p a /p > b /p
. If a > b. then we end up with 0 > 0. Can we divide both sides by the same nonzero quantity? Yes. If the quantity by which we multiply the statement through happens to be negative. then it is always true that b < a. we cannot reverse the order and still have a valid statement. then b > a. then a+c>b+d We can multiply both sides of a “strictly larger than” statement by the same positive quantity and get another valid statement. The “strictly larger than” relation turns into a “strictly smaller than” relation. However. It is never true that if a > b. Can we subtract the same quantity from both sides? Yes. c. For any numbers a. then for any positive number p ap > bp If the quantity by which we multiply through is 0.184 Equations and Inequalities
Here’s a summary of how we can morph “not equal” statements. and d. Can we multiply both sides by the same quantity? Only if that quantity is not 0. If a > b. If a > b. If a > b. then for any negative number n an < bn We can divide both sides of a “strictly larger than” statement by the same positive quantity and get another valid statement.

the sense of the inequality is reversed. If the quantity by which we divide through is negative. then for any negative number n a /n < b /n Here’s a summary of how we can morph “strictly larger than” statements. If a < b. then we end up with 0 < 0. If a > b. For any numbers a.Inequality Morphing
185
If the quantity by which we divide through is 0. If a < b. then a+c<b+d We can multiply both sides of a “strictly smaller than” statement by the same positive quantity and get another valid statement. • Can we reverse the order? Never. then b < a. then for any positive number p ap < bp If the quantity by which we multiply through is 0. then for any number c
a+c<b+c and a−c<b−c We can add two “strictly smaller than” statements (left-to-left and right-to-right) and get another “strictly smaller than” statement. just as with multiplication by a negative.
Manipulating < statements If some quantity a is strictly smaller than another quantity b. which is false. an > bn
. We can add or subtract the same quantity from each side of a “strictly smaller than” statement. If a < b. then for any negative number n. But we can reverse the order if we also reverse the sense of the inequality.” • Can we add the same quantity to both sides? Yes. we get undefined results on both sides of the inequality symbol. If a < b. if a < b and c < d. • Can we subtract the same quantity from both sides? Yes. then it is always true that b > a. the sense of the inequality is reversed. It is never true that if a < b. c. b. we cannot reverse the order and still have a valid statement. • Can we add one statement to another? Yes. • Can we multiply both sides by the same quantity? Only if that quantity is positive. • Can we divide both sides by the same quantity? Only if that quantity is positive. and d. unless we change the inequality to “strictly smaller than. If the quantity by which we multiply the statement through is negative.

For that reason. unless we change the inequality to “strictly larger than. then b ≥ a. if a ≥ b and c ≥ d. then it is always true that b ≤ a. We can add or subtract the same quantity from each side of a “larger than or equal” statement. then a+c≥b+d We can multiply both sides of a “larger than or equal” statement by the same nonnegative quantity and get another valid statement. Can we subtract the same quantity from both sides? Yes. If a ≥ b. then for any negative number n a /n > b /n Here’s a summary of how we can morph “strictly smaller than” statements. which is true. then for any number c
a+c≥b+c and a−c≥b−c We can always add two “larger than or equal” statements (left-to-left and right-to-right) and get another valid statement. If the quantity by which we divide through is negative. If a ≥ b. If a ≥ b. then for any positive number p a /p < b /p If the quantity by which we divide through is 0. then for any nonnegative number q aq ≥ bq If q = 0. the sense of the inequality is reversed. we end up with 0 ≥ 0. Can we multiply both sides by the same quantity? Only if that quantity is positive. If the quantity by which we multiply the statement is negative. we cannot reverse the order. But we can reverse the order if we also reverse the sense of the inequality. If a < b. • • • • • • Can we reverse the order? Never. we get undefined results on both sides of the inequality symbol. c. then for any negative number n an ≤ bn
. b. the sense of the inequality is reversed. except in the situation where a happens to equal b.” Can we add the same quantity to both sides? Yes. and d. Can we divide both sides by the same quantity? Only if that quantity is positive. Can we add one statement to another? Yes. it’s not generally true that if a ≥ b. For any numbers a.
Manipulating ê statements If some quantity a is larger than or equal to another quantity b.186 Equations and Inequalities
We can divide both sides of a “strictly smaller than” statement by the same positive quantity and get another valid statement. If a < b. If a ≥ b.

the sense of the inequality is reversed. then for any number c
a+c≤b+c and a−c≤b−c We can always add two “smaller than or equal” statements (left-to-left and right-to-right) and get another valid statement. But we can reverse the order if we also reverse the sense of the inequality.
Manipulating Ä statements If some quantity a is smaller than or equal to another quantity b. unless we change the inequality to “smaller than or equal. If a ≥ b. It’s not generally true that if a ≤ b. • Can we add one statement to another? Yes.” • Can we add the same quantity to both sides? Yes.Inequality Morphing
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We can divide both sides of a “larger than or equal” statement by the same positive quantity and get another valid statement. then for any negative number n a /n ≤ b /n Here’s a summary of how we can morph “larger than or equal” statements. If a ≥ b. • Can we reverse the order? Not in general. except when a happens to equal b. • Can we multiply both sides by the same quantity? Only if that quantity is nonnegative. then for any positive number p a /p ≥ b /p If the quantity by which we divide through is 0. If a ≤ b. If the quantity by which we divide through is negative. we cannot reverse the order. then a+c≤b+d We can multiply both sides of a “smaller than or equal” statement by the same nonnegative quantity and get another valid statement. c. • Can we subtract the same quantity from both sides? Yes. and d. For any numbers a. If a ≤ b. • Can we divide both sides by the same quantity? Only if that quantity is positive. then it is always true that b ≥ a. b. we get undefined results on both sides of the inequality symbol. then b ≤ a. We can add or subtract the same quantity from each side of a “smaller than or equal” statement. If a ≤ b. then for any nonnegative number q aq ≤ bq
. if a ≤ b and c ≤ d.

• Can we divide both sides by the same quantity? Only if that quantity is positive. which is true. the sense of the inequality is reversed. • Can we subtract the same quantity from both sides? Yes. If a ≤ b. you get 3 × (−1) < 7 × (−1) which simplifies to −3 < −7 This new statement is false! It becomes true if.188 Equations and Inequalities
If q = 0. • Can we multiply both sides by the same quantity? Only if that quantity is nonnegative. • Can we reverse the order? Not in general. If a ≤ b. If the quantity by which we multiply the statement is negative. then for any positive number p a /p ≤ b /p If the quantity by which we divide through is 0. then for any negative number n an ≥ bn We can divide both sides of a “smaller than or equal” statement by the same positive quantity and get another valid statement. then for any negative number n a /n ≥ b /n Here’s a summary of how we can morph “smaller than or equal” statements. Here’s an example: 3<7 If you multiply this through by −1 without changing the sense of the inequality. unless we change the inequality to “larger than or equal. you switch the inequality symbol from “strictly smaller than” to “strictly greater than” getting −3 > −7
. the sense of the inequality is reversed. we get undefined results on both sides of the inequality symbol. but only if.
Are you confused?
Some people find it hard to see why multiplying an inequality through by a negative number reverses its sense. If the quantity by which we divide through is negative. then we end up with 0 ≤ 0. If a ≤ b.” • Can we add the same quantity to both sides? Yes. • Can we add one statement to another? Yes.

whenever you multiply or divide an inequality through. It’s best to stick with plain numbers. or even equal to 0. the sense of the inequality must be reversed if the statement is to remain true. Check out the rules in this chapter to be sure you’re “obeying the law”! A single blunder can cause an error that may remain hidden for some time. Now you have something that contains an unknown.Inequality Morphing
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You must be careful if you multiply or divide an inequality through by a variable. Suppose you multiply the original inequality in this section through by x. also called constants. don’t ignore that uneasy feeling. Then you get 3x < 7x This changes the situation completely! You no longer have a plain statement of fact. that error will come around and bite you.
Here’s a challenge!
Suppose we are given the following inequality. First. the statement becomes false no matter what. let’s add 24 to each side. ask yourself.
Are you still confused?
A whimsical way to state the above warning is to invoke a time-worn proverb known as Murphy’s law: “If something can go wrong.” How can we do it?
Solution
We can use the rules for inequality morphing to change this statement into something with x on one side and a numeral on the other. it will. or by any expression containing a variable. If x happens to be negative. “Is this action completely safe? Can anything bad happen?” If you suspect possible trouble. But eventually. we must “solve the inequality.” Whenever you do anything to an equation or inequality. That gives us 34 < 2x − x which simplifies to 34 < x
. getting x + 10 + 24 < 2x − 24 + 24 This simplifies to x + 34 < 2x Now let’s subtract x from each side. If x happens to be 0. and we are told to derive a statement that clearly indicates all the real numbers x for which it is true: x + 10 < 2x − 24 In other words. Keep in mind that x might be positive or negative.

so we get a statement that says a positive integer is equal to itself ? 2. and a plain numeral all by itself on the right. We put the variable all by itself on the left-hand side of the relation symbol. Suppose we see this equation: 7/2 = 14/4 = 21/6 How can we simplify this using the rules for equation morphing.
Here’s a final challenge!
In terms of an inequality statement and set notation. by all means try it! 1. so we have x > 34 This is the standard way to state the solution to any single-variable algebra problem. How can we morph the equation in Prob. The solutions in the appendix may not represent the only way a problem can be figured out. Don’t hurry! You’ll find worked-out answers in App. 1 so we get a statement to the effect that a negative integer is equal to itself ? 3. although don’t know it at the time? 4. Therefore. B. If you think you can solve a particular problem in a quicker or better way than you see there. Here’s a hint: Use the same approach as we did in the final challenge. then x is larger than y. In logical form along with set notation. if x is an element of Z0+ and y is an element of R −. 5. Any negative real number we choose will be smaller than any nonnegative integer we choose. describe how the nonpositive real numbers relate to the nonnegative real numbers. In terms of an inequality statement and set notation. You may (and should) refer to the text as you solve these problems. Here’s a hint: Use the same approach as we did in the final challenge.
.190 Equations and Inequalities
We can say this in a more intuitive way by turning it around and reversing the sense of the inequality. In terms of an inequality statement and set notation.
Solution
Let’s call the set of nonnegative integers Z0+. and the set of negative reals R −. What happens if we multiply an equation through by the number 0? What happens if we multiply an equation through by a variable or expression that ultimately turns out to equal 0. we can write this as [(x ∈ Z0+) & (y ∈ R−)] ⇒ x > y
Practice Exercises
This is an open-book quiz. describe how the negative integers relate to the natural numbers. describe how the nonnegative integers relate to the negative real numbers.

For what real-number values of y is this inequality true? y /2 ≠ 4y + 7 10. In terms of an inequality statement and set notation. c) : [(a ≥ b) & (b ≤ c)] ⇒ (a = c) 8. For what real-number values of x is this equation true? x + 4 = 2x 9. Here’s a hint: Use the same approach as we did in the final challenge.Practice Exercises
191
6. For what real-number values of z is this inequality true? z /(−3) ≤ 6z + 6
. Write the following statement as a “mathematical verse. b.” Does it represent a valid mathematical law? If not. describe how the rational numbers relate to the irrational numbers. show a counterexample. (∀ a. 7.

Constants plus or minus x Here are some first-degree equations that contain a variable x with constants added and/or subtracted. Sums. place restrictions on them.
x−4=0 x + 7 = −2 a−x=0 a−5+x=0 a−x=b These equations can be morphed to get x all by itself on the left sides of the equality symbols. and Differences
193
In this equation. we can write a+x=b to represent an equation in x. The above equations then become: x=4 x = −9 x=a x=5−a x=a−b These are the solutions to the original equations. and as long as we know that x is the variable. Once we know that the values of a and b always stay the same no matter what happens to x.
. or c < e. Letters are used to avoid specifying exactly what numbers they are. We can use the rules from Chap. 9 to do this. d ≠ 0. So. and e doesn’t mean the exponential constant. b ≥ −5. we can subtract a from each side to get x=b−a Letter constants usually come from the first half of the English alphabet. for example. however. it’s easy. such as a > 1. we can morph the above equation to solve for x in terms of a and b. and nothing but constants on the right sides.Constants. We can. In this case. Letter constants are convenient when we want to show an equation in a certain form. All five constants are meant to stand for ordinary numbers. c does not stand for the speed of light. and are usually written in lowercase. Greek letters can also represent constants. because they clearly state the values of x in terms of the constants. as long as we realize that a and b represent constants.

Tables 12-1 through 12-5 show the processes in each case.
Statements x−4=0 x−4+4=0+4 x=4
Process for solving the equation x − 4 = 0. regardless of the sequence of steps. step-by-step in detail.
Reasons This is the equation we are given Subtract a from each side Simplify each side Add 5 to each side Simplify each side
Statements a−5+x=0 a−5+x−a=0−a −5 + x = −a −5 + x + 5 = −a + 5 x=5−a
. The processes shown here don’t necessarily represent the only ways the equations can be solved.
Process for solving the equation a − 5 + x = 0. You might want to try solving some of these equations in two or three different ways.194 First-Degree Equations in One Variable
Are you confused?
If you aren’t sure how the five solutions come out of the five original equations above. Nevertheless.
Reasons This is the equation we are given Add 4 to each side Simplify each side
Table 12-2.
Statements a−x=0 −x = −a −1(−x) = −1(−a) x=a
Process for solving the equation a − x = 0.
Reasons This is the equation we are given Subtract a from each side Multiply each side by −1 Simplify each side
Table 12-4. the solution to any equation should always turn out the same.
Table 12-1.
Statements x + 7 = −2 x + 7 − 7 = −2 − 7 x = −9
Process for solving the equation x + 7 = −2.
Reasons This is the equation we are given Subtract 7 from each side Simplify each side
Table 12-3.

so we have x = −x + (−b) + (−7) + c + a + (−5) We can add x to each side.Constants. x − a + 5 = −x − b − 7 + c
Solution
This can be done in various ways.
Statements a−x=b a−x+x=b+x a=b+x b+x=a b+x−b=a−b x=a−b
Process for solving the equation a − x = b. This removes another constant from the left side. That gets rid of the variable on the right side. and again simplify the left side. let’s change all the subtractions to negative additions before we start rearranging things. Now we have 2x = −b + (−7) + c + a + (−5) Let’s rearrange the right side to get the letter constants in alphabetical order. followed by the numerals. and an expression containing the constants without x on the right side. but it will make things more elegant in the end. (That’s not technically necessary. Sums. to get x = [a + (−b) + c + (−12)]/2
.
Reasons This is the equation we are given Add x to each side Simplify the left side Transpose the left and right sides Subtract b from each side Simplify the left side
Here’s a challenge!
Manipulate the following equation so it contains x all by itself on the left side.) That gives us 2x = a + (−b) + c + (−7) + (−5) We can divide through by 2. This gets one of the constants out of the left side. and then simplify the left side. and then simplify both sides. They’ll all produce the same result. leaving only constants there. and then add the two plain numbers in the numerator on the right-hand side. To avoid making errors with the signs. so we have x + 5 = −x + (−b) + (−7) + c + a Next. That gives us x + (−a) + 5 = −x + (−b) + (−7) + c We can add a to each side. let’s add −5 to each side. and Differences
195
Table 12-5.

This gives us x = a /2 − b /2 + c /2 − 6
Products and Ratios
Let’s see what happens when the quantities on either side of an equation are multiplied by constants.196 First-Degree Equations in One Variable
The right-hand distributive law for division over addition. Also. in the fourth solution equation (and in Table 12-9). applied to the expression on the right side of the equation. Note that in the third and fifth original equations above (and in Tables 12-8 and 12-10). 9. Here are the results. step-by-step.
4x = 0 x/7 = 2 2x /a = b 5abx = c 3x /(4a) = 3 Using the rules from Chap. or both. gives us x = a /2 + (−b)/2 + c /2 + (−12)/2 We can simplify the right-hand side by changing all the negative additions back to subtractions. and the constants all by themselves on the right. x=0 x = 14 x = ab /2 x = c /(5ab) x = 4a
Are you confused?
If you can’t see straightaway how these solutions are derived. we can manipulate these equations to get x alone on the left side. divided by nonzero constants. Tables 12-6 through 12-10 show how the equations can be solved.
Examples Here are five first-degree equations that contain a variable x multiplied and/or divided by constants. That solves the equations.
. we must not let a equal 0. we must never allow either a or b to equal 0. and then dividing out the numeral quotient.

Statements 2x /a = b a(2x /a) = ab 2x = ab 2x /2 = ab /2 x = ab /2
Process for solving the equation 2x /a = b.
Reasons This is the equation we are given This will allow us to solve the equation in a “streamlined” fashion Multiply through by the constant (4a) Simplify both sides Divide each side by 3 Simplify each side
Statements 3x /(4a) = 3 Consider (4a) to be a single constant [3x /(4a)](4a) = 3 × (4a) 3x = 12a (3x)/3 = 12a /3 x = 4a
.
Statements 4x = 0 4x /4 = 0/4 x=0
Process for solving the equation 4x = 0.
Process for solving the equation 3x /(4a) = 3.
Reasons This is the equation we are given Divide each side through by 4 Simplify each side
Table 12-7. provided a ≠ 0.
Reasons This is the equation we are given Multiply each side by a Simplify the left side Divide each side by 2 Simplify the left side
Table 12-9.
Reasons This is the equation we are given We are about to divide through by the product of these constants This will allow us to solve the equation in a “streamlined” fashion Divide through by the constant (5ab) Simplify the left side
Statements 5abx = c Require that a ≠ 0 and b ≠ 0 Consider (5ab) to be a single constant 5abx /(5ab) = c /(5ab) x = c /(5ab)
Table 12-10.
Reasons This is the equation we are given Multiply each side by 7 Simplify each side
Table 12-8. provided a ≠ 0 and b ≠ 0.Products and Ratios
197
Table 12-6.
Process for solving the equation 5abx = c. provided a ≠ 0.
Statements x /7 = 2 7x /7 = 7 × 2 x = 14
Process for solving the equation x / 7 = 2.

When we do this. constants can be added to or subtracted from that variable. the left-hand side of the equation becomes undefined.
Examples Here are some first-degree equations that involve combinations of sums. The variable. Let’s multiply the equation through by the quantity (4cd).
Combinations of Operations
In a first-degree equation that involves a single variable. x. differences. and the variable can also be multiplied or divided by nonzero constants. we must insist that a ≠ 0 and b ≠ 0. which constants cannot equal 0. and an expression containing the constants without x on the right side. so the equation is a first-degree equation. Indicate. 3abx /(4cd ) = k 2
Solution
We must have c ≠ 0 and d ≠ 0 because. if applicable.198 First-Degree Equations in One Variable
Here’s a challenge!
Manipulate the following equation so it contains x all by itself on the left side. The square of a constant is always another constant. That produces 3abx /(3ab) = 4cdk 2/(3ab) which simplifies to x = 4cdk 2/(3ab) That does it! We don’t have to worry about the fact that one of the constants is squared. products. if either of them are allowed to equal 0. and ratios:
8x − 4 = 0 18x + 7 = −2 a − 3x = 0 a − 5 + 15x = 0
. is never raised to any power (other than the first power). We get [3abx /(4cd )](4cd ) = (4cd )k 2 which simplifies to 3abx = 4cdk 2 Now we can divide the entire equation through by the quantity (3ab).

Table 12-11. putting fractions in lowest terms. so that x appears alone on the left sides of the equality symbols.Combinations of Operations
199
a − 8x = b −6a + 3x = 12b 6a − 3x /(bc) = −24d These seven equations can all be rearranged with the morphing laws we already know. Some of the steps are combined. and getting the letters for the constants in alphabetical order. and nothing but constants appear on the right sides.
Statements 18x + 7 = −2 18x = −9 x = −9/18 x = −1/2 Reasons This is the equation we are given Subtract 7 from each side Divide each side by 18 Put the fraction into lowest terms
. it’s necessary that b ≠ 0 and c ≠ 0. Also note that an attempt has been made to put the solutions in elegant form by avoiding sums or differences in the numerators of fractions. Streamlined process for solving the equation 18x + 7 = −2.
Statements 8x − 4 = 0 8x = 4 x = 4/8 x = 1/2 Reasons This is the equation we are given Add 4 to each side Divide each side by 8 Put the fraction into lowest terms
Table 12-12. making the derivations less tedious than those earlier in this chapter. Here are the respective solutions: x = 1/2 x = −1/2 x = a /3 x = 1/3 − a /15 x = a /8 − b /8 x = 2a + 4b x = 2abc + 8bcd
Are you confused?
Tables 12-11 through 12-17 break down the solution processes for the above equations. Streamlined process for solving the equation 8x − 4 = 0. Note that in the last original equation above (and in Table 12-17).

200 First-Degree Equations in One Variable
Table 12-13.
Statements a − 8x = b −8x = b − a −x = (b − a)/8 x = −(b − a)/8 x = (a − b)/8 x = a /8 − b/8 Reasons This is the equation we are given Subtract a from each side Divide through by 8 Multiply through by −1 Simplify the right side Right-hand distributive law for division over subtraction
Table 12-16. change negative addition to subtraction Divide through by 15 Right-hand distributive law for division over subtraction Reduce the numerical fraction to lowest terms
Table 12-15. Streamlined process for solving the equation −6a + 3x = 12b.
Statements −6a + 3x = 12b 3x = 6a + 12b x = (6a + 12b)/3 x = 2a + 4b Reasons This is the equation we are given Add 6a to each side Divide through by 3 Right-hand distributive law for division over addition
. Streamlined process for solving the equation a − 5 + 15x = 0.
Statements a − 3x = 0 −3x = −a 3x = a x = a /3 Reasons This is the equation we are given Subtract a from each side Multiply through by −1 Divide through by 3
Table 12-14. Streamlined process for solving the equation a − 3x = 0.
Statements a − 5 + 15x = 0 −5 + 15x = −a 15x = −a + 5 15x = 5 − a x = (5 − a)/15 x = 5/15 − a /15 x = 1/3 − a /15 Reasons This is the equation we are given Subtract a from each side Add 5 to each side Commutative law for addition. Streamlined process for solving the equation a − 8x = b.

This is called the standard form for a firstdegree equation in one variable. then x disappears. It’s possible that b can equal 0. and it’s not a first-degree equation in one variable—because there is no variable! Here are several examples: x=0 3x = 0 −4x = 0 x+3=0 x−4=0 5x + 2 = 0 5x − 2 = 0 −5x + 2 = 0 −5x − 2 = 0
Here’s a challenge!
Imagine that you are working on a problem in physics.
Statements 6a − 3x /(bc) = −24d −3x /(bc) = −6a − 24d 3x /(bc) = 6a + 24d 3x = (6a + 24d )(bc) 3x = 6abc + 24dbc 3x = 6abc + 24bcd x = (6abc + 24bcd )/3 x = 2abc + 8bcd Reasons This is the equation we are given Subtract 6a from each side Multiply through by −1 Multiply through by (bc) Right-hand distributive law for multiplication over addition Commutative law for multiplication in second addend on right side Divide through by 3 Right-hand distributive law for division over addition
Standard form Whenever you encounter an equation that can be morphed into the following form. engineering. or some other branch of science and you come across this equation:
4/x − 8/k = 0
. Streamlined process for solving the equation 6a − 3x /(bc) = −24d. the statement ends up trivial or false.Combinations of Operations
201
Table 12-17. and a and b are constants. If a = 0. then that equation is a first-degree equation:
ax + b = 0 where x is the variable. provided b ≠ 0 and c ≠ 0.

and can be treated as one. how? If not. how? If not. because in the equation as stated. At first glance. getting 4 = 0x
. Therefore. Note the similarity of this to the equation in the previous challenge: 4/x = 0 Can this be rearranged into standard first-degree equation form? If so. However. it appears in the denominator of a fraction. But if a singlevariable equation can be converted into standard first-degree form. is raised to the −1 power. getting 4/x − 8/k + 8/k = 8/k which simplifies to 4/x = 8/k Using the rule of cross-multiplication. Can this be rearranged into standard first-degree equation form? If so. it might seem that the original equation can’t be first-degree. why not? What is the significance of the result?
Solution
Because k is a constant. the original equation is a first-degree equation. getting −8x + 4k = −8x + 8x which simplifies to −8x + 4k = 0 This is in standard first-degree form. With that in mind. which is the denominator of a fraction. why not? What is the significance of the result?
Solution
You can rewrite this as 4/x = 0/1 Then you can try using the rule of cross-multiplication. you can add the quantity (8/k) to each side of the equation. if you let a = −8 and b = 4k. the quantity (8/k) must also be a constant. ax + b = 0. because it looks as if x. then it is a first-degree equation.
Here’s another challenge!
Suppose you come across this equation in a physics or engineering problem. k cannot be allowed to equal 0.202 First-Degree Equations in One Variable
where k is a constant. you know that 4k = 8x Now you can add −8x to each side.

Let’s call our unknown number x. Problems that seem simple to us were difficult for them. and then add the result to 1/5 of itself. even to the best mathematicians. What is the original number? Solution A The first step in solving any word problem is to set up an equation representing that problem. so the variable x becomes meaningless. getting 0x − 4 = 0
203
It’s tempting to think that this is in standard form for a first-degree equation.
Problem A Imagine a number.
Word Problems
Centuries ago. The equation is
x + x /2 + x /5 = 51/10 We can multiply through by 10 to get 10x + 5x + 2x = 51 Let’s apply the right-hand distributive law for multiplication over addition “in reverse” on the left side of the equation. they often sought out solutions by making educated guesses until they “got lucky. We add it to half of itself. No matter what real number you choose for x. we get 51/10. and then add x /5 to that. When they encountered word problems like the ones that follow. But the constant by which you multiply x (called the coefficient of x) is 0.Word Problems
If you reverse the sense of the equation. you have 0x = 4 Now you can subtract 4 from each side. the algebra in this chapter was unknown.” We have a better way. then add x /2 to that. That gives us (10 + 5 + 2)x = 51 which simplifies to 17x = 51 We can divide through by 17 and get x=3
. you get an absurd result. The statement of the problem tells us that if we take x. The final sum is equal to 51/10.

500.000 If we multiply through by 3/4. Jack received $37.500.204 First-Degree Equations in One Variable
Problem B The old Widow Johnson sold all her property and put the cash into a savings account. in dollars. Jane has two children to support now. The account contained $150.000
.500. like her mother. or $112. The total inheritance. We can also figure Jane’s share by noting that she got $150. decided that Jane ought to get three times as much of the inheritance as Jack. When we set up the equation on this basis. Therefore. She left two children: Jane and Jack. then Jack’s share was y /3. we get
y + y /3 = 150.500. or $112. Jane got married and then. How much did each child get? Solution B Let’s call Jack’s share of the inheritance. became a widow. in dollars.500. x. Solution C If we call Jane’s share of the inheritance y. in dollars.000 when she died.000 which can be simplified to (4/3)y = 150. we get (3/4)(4/3)y = 3/4 × 150. rather than Jack’s share.000 We can simplify this to 4x = 150.
Problem C Solve the previous problem by letting y represent Jane’s share of the inheritance.
x + 3x = 150.000 Then we divide through by 4 to get x = 37. Then Jane got 3x. was 150. Knowing that Jane would likely need more money than Jack. in the wisdom of her waning weeks.000. the old Widow Johnson.500 Therefore. in dollars. That was 3 × $37.000 − $37. and Jack is a bachelor living alone. Jane got the other portion of the inheritance.

000 when he died.000.000 = 130.000 less than Jill.000 less than Judy.500.000) + (x + 30.000 dollars. or $37.500. we obtain x + x + x + 10. in the dullness of his demise.000 from each side gives us 3x = 90.000.000 more than Jill. Judy got $20. that Joann received. We have
x + (x + 10. that Joann should get $10.500. or $37.000) = 130.000 Subtracting 40. and Jill should get $20. the old Widower Jones decreed. Jill. Nevertheless.
Problem D The old Widower Jones (who was a good friend of the old Widow Johnson) sold all his property and put the money into a savings account.000 + 30.000.000 That means Joann received $30. Jill got $10. Jack got the other portion. and Judy. We can also figure Jack’s share by noting that he got $150.000) + 20. The account balance was $130.500.000 Ungrouping these addends and then rearranging them according to the commutative law for addition.000 Dividing through by 3.000 dollars.Word Problems
205
which simplifies to y = 112. Judy got (x + 10. Then Jill got x + 10. He left three children: Joann.000 more than Joann.500. we get x = 30.000 − $112. How much did each child get? Solution D Let’s say that x was the amount. in dollars. All three are married today. or x + 30. or $40.000 dollars. and well-off.000 = 130. which was 1/3 of $112.500 Jane’s share was therefore $112.
.000 which simplifies to 3x + 40. or $60.

Add 60 to the previous result. You’ll get this sequence of operations: • • • • • • • Start with 10. you’ll be able to get the equation into the form ax + b = 0 where a and b are constants. Choose any real number you want. This example is only one of infinitely many variations on the same theme. Subtract 10 from the previous result. Perform the following operations in this order: • • • • • • • Multiply the number you have chosen by 2. this might not be possible. every subtraction with addition.
Explain how this trick works. However.206 First-Degree Equations in One Variable
Are you confused?
Suppose you’re trying to solve a word problem. getting 2x. and then perform multiple operations on successive results to obtain x. If you do that. getting x. Add 10 to the previous result. replacing every addition with subtraction. Divide the previous result by 2. Subtract the number you originally picked. You will end up with 10. and you set up an equation in one variable to represent it. Add x.” You pick the number that you want the game to end up at (in this case 10). Subtract 60 from the previous result. getting 40 + 4x. no matter what number you originally chose. you’ll have to use more powerful equation-solving techniques than those in this chapter.
Solution
Games like this are fabricated by “reverse engineering. obtaining 10 + x. You can see how this example was put together by going through the above sequence of steps backward. You’ll learn about them in Part 3. Multiply the previous result by 2.
Here’s a challenge!
Try playing the following little number game.
When you build games of this sort. You hope to get a first-degree equation.
. and every division with multiplication. If you’re lucky. Divide the previous result by 4. Divide the previous result by 2. Multiply the previous result by 4. you must never multiply or divide by any expression containing x. you make it possible for someone to defeat the trick by choosing x such that division by 0 occurs somewhere in the sequence of steps. If you can’t get the equation into the standard form for a first-degree equation. every multiplication with division. getting −20 + 4x. getting −10 + 2x.

you get −1. Find the number by devising a first-degree equation in x. Solve. Bill weighs 10 kilograms (kg) more than Bruce. by all means try it! 1. Using the narration method. and end up with 135. and divide that sum by 4. and our boat’s water speed is always 18 mi/h. and then solving for x.
. slowed us down. which we were fighting. morph this into standard form for a first-degree equation in one variable: x − 7 = 7x + x /7 4. Don’t hurry! You’ll find worked-out answers in App.Practice Exercises
207
Practice Exercises
This is an open-book quiz. How fast was the river flowing? Assume that the river flowed at the same speed all during our journey. in no more than two steps each. Using the S/R table method. 6. Using the S/R table method. then add 8 to that product. The trip takes 1 hour and 12 minutes (1 h 12 min). We would expect it to take exactly 1 h if there were no current in the river. What does each person weigh? 9. we divide the result by 2. But the current. If you think you can solve a particular problem in a quicker or better way than you see there. We make a trip upstream from our cabin to our cousins’ cabin. morph this into standard form for a first-degree equation in one variable: 4x + 4 = 2x − 2 2. Suppose we take a certain number. 1 through 4. Assuming the river keeps flowing at the same speed as we determined when we solved Prob. and then subtract 1/10 of itself. The solutions in the appendix may not represent the only way a problem can be figured out. You may (and should) refer to the text as you solve these problems. running the boat at top speed all the way. After that. Using the S/R table method. how long will it take us to travel downstream from our cousins’ cabin back to our own? Here’s a hint: The travel time (in hours) equals the distance traveled (in miles) divided by the constant speed (in miles per hour). and then solving for x. Imagine that we have a motorboat with a maximum water speed (the speed relative to the water it’s floating on) of 18 miles per hour (mi/h). morph this into standard form for a first-degree equation in one variable: x /3 = 6x + 2 3. and Bruce weighs 5 kg more than Bonnie. a distance of 18 miles (mi). B. When you multiply a certain number by 2. the standard-form equations you derived in Probs. The combined weight of all three people is 200 kg. 10. 8. morph this into standard form for a first-degree equation in one variable: x /3 + x /6 = 12 5. 7. 9. Find the number by devising a first-degree equation in x.

the domain and the maximal domain of a mapping are identical. The range is a subset of the co-domain. Similarly. of the points inside the maximal domain and the co-domain are actively involved in the mapping. Sometimes. 13-1. 13-1 contain more points than those inside the ovals.
Maximal domain and co-domain The large rectangles in Fig. The top rectangle is called the maximal domain of the mapping. The co-domain is the set of all people in the lower van. If that were the case in Fig. In our example. the codomain and the range of a mapping might be identical. you can define the mapping in terms of ordered pairs. The domain is a subset of the maximal domain. Ordered pairs When you have a mapping from the elements of one set to the elements of another set. but not all. If that were true in Fig. 13-1. Some. the maximal domain of the mapping is the set of all people in the upper van. and the bottom rectangle is called the co-domain. whether they’re sending messages or not.Mapping “Territories”
209
c a e b d f
Domain Maximal domain
y x w v Co-domain
z
Range
Figure 13-1 A mapping between two sets. then the hatched region would fill the top rectangle. whether they’re receiving messages or not. then the hatched region would fill the bottom rectangle. An ordered pair is an expression in parentheses
.

In the lower van. (c. Also imagine that the upper oval represents
Positive rationals Positive reals
Mapping: Multiply by –1
Negative reals Negative rationals
Figure 13-2 A mapping from the positive rational
numbers to the negative rational numbers. 13-1.
Here’s a challenge!
Examine Fig. Relations.y). as you’ll see later in this chapter. y. you can (but don't have to) put a space after the comma. (c. Points v. In the situation shown by Fig. The second element represents an element of the range. Point c in the domain maps to three points in the range.z). 13-1. (e. When you write an ordered pair. and the lower rectangle represents the set of all negative real numbers. and z in the range are each mapped from two points in the domain.x). the correspondence between the points in the domain and the points in the range is not one-to-one. (c. and (f.
. and Functions
that contains two items separated by a comma. one person is sending messages to three different people in the lower van.v). In some situations. (d. You can imagine that in the upper van. Suppose the upper rectangle represents the set of all positive real numbers.
Are you confused?
In the example of Fig. The first item represents an element of the domain.v). “dupes” are not allowed.w). as you would in an ordinary sequence or a list of set elements.z).y). “Dupes” like this are okay in a general mapping. three people are receiving messages from two different senders. 13-2. the ordered pairs are (a.210 Mappings. (b.

and every element y maps from a single element x.
Injection Figure 13-3 shows a situation in which elements of set X are mapped to elements of set Y. It also fails to tell us how we would map anything to a positive real. Here’s the mapping: any number x in the upper oval is mapped into a number y in the lower oval by taking its additive inverse (multiplying x by −1). The range is the set of all negative rationals.
Types of Mappings
Mathematicians have special names for different types of mappings. even though they may seem strange at first! Imagine two sets of objects. This mapping has a domain that is a subset of X. The domain is the set of all positive rationals. The maximal domain is the set of all positive reals. There are three major ways in which the elements of X can be mapped to the elements of Y. and the lower oval represents the set of all negative rationals. and let the variable y represent an element in set Y. called set X and set Y. The co-domain is the set of all negative reals. You should know what these terms mean. where x is rational and x > 0.Types of Mappings
211
the set of all positive rationals.−x). and a range that is a subset of Y. This mapping does not tell us how to map a negative real number to anything. Let the variable x represent an element in set X.
. Each element x
Set Y = all possible values of variable y Range
Domain
Set X = all possible values of variable x
Figure 13-3 An example of an injection. How can we define the ordered pairs in this mapping? What is the domain? The maximal domain? The range? The co-domain? What happens in this situation if we want to map a negative real number to something. Every element
x maps into a single element y.y) as always having the form (x. or if we want to map something to a positive real number?
Solution
We can define the ordered pairs (x.

Every possible
element y is accounted for. “A bijection is both one-to-one and onto.” which some people call an injection. because it sounds too much like “one-to-one. Here. Relations. a surjection is sometimes called an onto mapping. called a bijection. or simply onto. That term is rarely used nowadays. A mapping of this type is called an injection. This type of mapping is called a surjection. The domain is a subset of X. But that’s a little misleading. and is the result of a mapping from at least one element x. 13-4 is a little different. Another expression you might hear is. You may occasionally hear an injection called a one-to-one mapping.” The old-fashioned term for a bijection is one-to-one correspondence. it clearly isn’t! Bijection Figure 13-5 shows an example of a third type of mapping.
in set X corresponds to one. This is an injection that is also a surjection. but the range is identical to Y.
. between two sets X and Y. elements of X are mapped to all the elements of Y. or simply one-to-one.
Surjection The mapping in Fig. and Functions Range = set Y = all possible values of variable y
Domain
Set X = all possible values of variable x
Figure 13-4 An example of a surjection. but it doesn’t have to be. because an injection doesn’t necessarily involve all the elements of either set X or set Y. A surjection can be one-to-one. element y in set Y. but only one.212 Mappings. and in this example. Because it maps elements of set X completely onto set Y.

the domain is shown as a proper subset of set X.
Are you confused?
In Fig. 13-4. which is a special sort of injection!) In Fig. Give an example of a bijection between X and Y. Again. or both. we get a number y in set Y that’s larger than 1 but smaller than 2. and the range is shown as a proper subset of Y.
Here’s a challenge!
Let X be the set of all real numbers x larger than 0 but smaller than 1. the domain is shown as a proper subset of X. (If both were true. Give an example of an injection from X into Y.Types of Mappings Range = set Y = all possible values of variable y For every x. Let Y be the set of all real numbers y strictly larger than 1. there is exactly one x
Domain = set X = all possible values of variable x
Figure 13-5 An example of a bijection. the domain could contain all of the elements in set X. we’d have a bijection.
Solution
If we add 1 to any number x in set X. We can write y=x+1
. or the range could be the entire set Y. 13-3. there is exactly one y
213
For every y. the domain could be the entire set X. However. It is both an injection
and a surjection. but the range is shown as the entire set Y.

or exactly one) real number y larger than 1. shown in Fig. we can always find a unique real number x between 0 and 1 that has y as its reciprocal. 0 and 1 to the set Y of all reals strictly larger than 1.
Figure 13-7 A bijection between
the set X of all reals between 0 and 1 and the set Y of all reals strictly larger than 1. Open circles indicate points not included in the domain and range. but not including. is a bijection. and vice-versa. Conversely. no matter what real number y larger than 1 we choose. Open circles indicate points not in the domain and range. 13-6. Now let’s consider a different mapping. and Functions
Figure 13-6 An injection from
the set X of all reals between. 5 4 y=x+1 3 2 1 Set X 0 0 3 Set Y 2 1 5 4
This mapping is an injection. and the range happens to be a proper subset of Y. This mapping. because for any x in the domain.214 Mappings. because every element in set Y is accounted for. as shown in Fig. The domain is the whole set X. which are shown as heavy gray lines. But it’s not onto the entire set Y. which are shown as heavy gray lines. we get a number y in set Y that’s larger than 1. No matter what x between 0 and 1 we choose. there’s a single y in the range. If we take the reciprocal of any number x in set X. We can write y = 1/x This mapping is a bijection. the reciprocal is always a unique (one and only one. 5 4 y = 1/x 3 2 1 Set X 0 0 3 Set Y 2 1 5 4
. 13-7. This mapping is one-to-one. Relations.

Examples of Relations
215
Examples of Relations
Whenever you can express a mapping in terms of ordered pairs.−3) If x = −1. then (x. Ordered pairs are produced by putting values for x into the equation. the elements of the domain and the range can be represented by variables. By definition. Let the range be the set of all real numbers y. We can also call x the “input variable” and y the “output variable.y) = (4.
A surjective relation Suppose both the maximal domain X and the co-domain Y of a particular mapping include all real numbers. Let the essential domain be the set of all nonnegative real numbers. We write the dependent variable all by itself on the left side of the equality symbol.2) If x = 3/2. then (x. there is exactly one value of y.
Independent vs. A relation therefore maps values of the independent variable to values of the dependent variable. because for every value of x. then that mapping is a relation. When we put specific values of x into this. therefore. and vice-versa.y) = (3/2.27)
This mapping is one-to-one. the mapping is an injection. This section will give you some more examples.6) If x = 25.1) If x = 0.y) = (0. then (x. then x is the independent variable and y is the dependent variable.y) = (−1. then (x. If we say that x is a nonspecific element of the domain and y is a nonspecific element of the range.y) = (−5. We can call this relation an injective relation. then (x. and then calculating the values for y. and then write an expression containing the independent variable on the right side.7/2) If x = 4. An injective relation Relations between sets of numbers are often represented by equations. dependent variable In a relation.y) = (25. the set of all x such that x ≥ 0. Here is an example:
y=x+2 for all real numbers x. then (x.” as computer scientists sometimes do. so it is the same as the co-domain. that is. Now consider this equation:
y = ±(x1/2)
. we get results such as: • • • • • • If x = −5. The examples in the “challenges” you’ve seen so far in this chapter are all relations.

then (x. then (x.y) = (4. we can square it and get a nonnegative real number x.y) = (4.−1) If x = 4.y) = (9. That means the mapping is a surjection.y) = (1/9. there are two values of y. then (x. then (x.y) = (0. then (x.0)
We now have a one-to-one mapping. No matter what nonnegative real number x we plug into this relation.−1/2) If x = 1. It also works the opposite way: No matter what nonnegative real y we want to get out of this relation.y) = (9.1) If x = 4. No matter what real number y we choose. we get results such as: • • • • • • If x = 1/9. and Functions
When we plug specific values of the independent variable x into this equation.1/2) or (1/4. then (x.3) or (9.2) If x = 9.3) If x = 0. then (x.1) or (1. then (x.−3) If x = 0.−2) If x = 9. Then we get this equation to represent it:
y = x1/2 When there is no sign in front of an expression raised to the 1/2 power. then (x. But the mapping is onto the entire co-domain.−1/3) If x = 1/4.y) = (1. We’ve simply declared that all negative output values are invalid! Here are some of the ordered pairs in this relation: • • • • • • If x = 1/9. then (x.
A bijective relation Let’s modify the relation in the preceding section by restricting the co-domain and range Y to the set of nonnegative reals. then (x. That means it’s both injective and surjective.y) = (1.216 Mappings. Relations.
. so we can call this relation a surjective relation. The equation y = x1/2 represents a bijective relation within the set of nonnegative reals. we can find a unique nonnegative real x to plug in that will give it to us. the 1/2 power indicates the nonnegative square root.1/3) If x = 1/4. Now there’s only one output value y for every input value x. then by convention.1/2) If x = 1. we get a unique nonnegative real number y out of it.1/3) or (1/9. then (x.y) = (1/4.2) or (4. and it’s also onto the entire co-domain.y) = (1/9.y) = (1/4.0)
This mapping is clearly not an injection! For every nonzero value of x.y) = (0.

and you’ll see how the above relations look when illustrated that way. Is this relation injective? Is it surjective? Is it bijective?
Solution
Let’s represent natural numbers by the independent variable n. 9) suggests a way to define a relation between the set of naturals and the set of rationals. and map it into the rational number 0 for r.Examples of Relations
217
Are you confused?
Graphs can make relations easier to understand.
. List the first 13 ordered pairs in this relation. We can start with 0 for n. Then we
Figure 13-8 This two-dimensional list of all rationals (duplicated
from Chap. you’ll learn about one of the most common graphing schemes in mathematics. 13-8 as a guide.
Here’s a challenge!
Using Fig. define a relation that maps the set of natural numbers onto the set of rational numbers. In the next chapter. and corresponding rational numbers by the dependent variable r.

(4. (2. 13-8 along the dashed “square spiral. there can never be more than one value for the dependent variable. because there’s never more than one value of y for any value of x. for every value of the independent variable. (6.”
Square the input Now let’s look at another simple relation. the dependent variable is replaced by the function letter followed by the independent variable in parentheses.−1) and (11. (10. Mathematicians name functions by giving them letters of the alphabet such as f.218 Mappings. But when r = 2/2 (which is equal to 1). Let’s look at some examples of functions.−2/2) have the same value of r.” increasing n by 1 each time we move to the next block. we have n = 7. (1. For example.1/2). g. and the range is the set of nonnegative reals.−1/2). “f of x equals x plus 1. because −1 = −2/2. (11. If we keep writing out the list of ordered pairs for a long time. But it’s a surjection. and it is one-to-one. and h. and letting r be equal to the rational number in the block. this relation can’t be a bijection. The domain is the entire set of reals. a single value of the dependent variable might be mapped from two. For every rational number r. (5. The same thing happens with many other rationals.
Examples of Functions
A function is a relation in which every element in the domain has at most one element in the range. (7. Relations.
Add 1 to the input Consider a relation in which the independent variable is x and the dependent variable is y. but the relation we’ve defined here is not of that sort! Because it’s not an injection. We might write f (x) = x + 1 to represent the above equation. the ordered pairs (4. 9. (8. and for which the domain and range are both the entire set of real numbers. For example.2/3). or more values of the independent variable—even infinitely many. the ordered pairs in the form (n. for every value of x. In this notation.2/2). then n = 2. four.−2/3).−2). An injection must be one-to-one. (12. and so on forever! Some of the r values in the range of this relation have more than one n counterpart in the domain.1). if r = 1. Our relation is defined as follows:
y=x+1 This is a function between x and y. (9. we can always find at least one natural number n that maps to it. (3. In fact. This function is bijective. Remember that in Chap. It maps values of x onto the entire set R. and then we can say. three.−1).1/3).2). we deliberately engineered this two-dimensional list so it would account for every possible rational number. and Functions
can proceed in Fig. When we follow that pattern. we’ll keep coming across “dupes” such as this.−1/3).r) are (0. Here’s the equation that represents the relation:
w = v2
. The independent variable is v and the dependent variable is w.−2/2).0). there is exactly one value of y. That is. Even so.

it cannot be bijective. For every nonzero value of w in the range of g. It maps onto the entire range. Let’s call the independent variable t and the dependent variable u. These two values are additive inverses (negatives) of each other. If we call it g. because every possible value in its range (the set of nonnegative reals) is accounted for. however. there is exactly one value of w. Which one?
. Our function g is not injective because it’s two-to-one. you can transpose the values of the independent and dependent variables while leaving their names the same. If we call it h. so it’s injective. For every value of u in the range of h. That means h is a bijection. its inverse is written h−1(z). then h(t) = t3 For every value of t in the domain of h. Then it can’t be a function. there is exactly one t in the domain. The domain and range are both the entire set of reals. The inverse of a function. g. we can write g(v) = v2 For every value of v in the domain of g. When you do these things.Examples of Functions
219
This is another example of a function. This function is one-to-one. If you have h(z). The trouble happens when a relation is one-to-many. you get another relation. The inverse of one of these functions is not a function. then v = 7 or v = −7. But the reverse is not true.”
Cube the input Here’s another relation. so it’s surjective as well. graphing can be useful when you want to see how functions map values of an independent variable into values of a dependent variable. which we can also call g(v). is not necessarily another function! Look again at the three functions f. The inverse of a relation is denoted by writing a superscript −1 after the name of the relation. there is at least one v in the domain such that g(v) = w. The function g is surjective. The reverse is also true. For example. if w = 49. not one-to-one (except when v = 0).
Here’s a challenge!
With any relation. In formal language we say. which is called the inverse relation (or simply the inverse if the context is clear). The equation is
u = t3 This is a function.
Are you confused?
As with relations. and h above. Graphs of the above three functions are shown in the next chapter. Therefore. a relation can be many-to-one and still be a function. however. there is exactly one value of u in the range. You can also transpose the domain and the range. “For any nonnegative real number w in the range of g. in the range. there are two values of v in the domain. This is not a problem.

We have f (x ) = x + 1 and f –1(x ) = x − 1 We also have h (t ) = t 3 and h −1(t) = t1/3 If a function is one-to-one over a certain domain and range.
. and is therefore not a function. But the inverse is one-to-two (except when w = 0). Remember that the domain is the entire set of reals. and that’s okay. You can also write g−1(v) = ±(v 1/2) The function g is two-to-one (except when v = 0). and the range is the set of nonnegative reals: w = g (v) = v2 If you take the equation w = v2 and transpose the positions of the independent and dependent variables. you’ll get two values of w. so their inverses must also be one-to-one. Both f and h are one-to-one. you get v = w2 This is the same as w = ±(v 1/2) The plus-or-minus symbol is important here! It indicates that for every value of the independent variable v you plug into this relation. and you can transpose the domain and range. but prevents it from being a function. Here’s the function again. If a function is many-to-one. That makes g−1 a legitimate relation.220 Mappings. one positive and the other negative. then you can transpose the values of the independent and dependent variables while leaving their names the same. The other two functions above have inverses that are also functions. The resulting inverse is a function. and Functions
Solution
The inverse of g is not a function. then its inverse is one-tomany. Relations.

) 5. (No two boys have the same name. we find the reciprocal of
. what type of relation is this? Is it injective? Is it surjective? Is it bijective? 7. What sort of mapping did we carry out from set A to set B ? What was the maximal domain? What was the essential domain? What was the co-domain? What was the range? (Note: As a result of our behavior. Imagine a dance at which there are 15 boys and 15 girls. and the essential domain is Z.000 members. what type of mapping is it? 3. so all the slips get taken.) The girls pick slips out of the jar. If we think of this action as a mapping from the set of boys to the set of girls. Every boy writes his name on a slip of paper. which had 60. Consider the set Q of all rational numbers and the set R of all real numbers. but the other five have to contend with two girls! If we think of this action as a mapping from the set of boys to the set of girls. to determine who their dance partner will be. and let B be the set of all members of Beta. Every single member of Beta got our message. Don’t hurry! You’ll find worked-out answers in App. Every girl writes her name on a slip of paper. Imagine another dance at which there are 10 boys and 15 girls. Suppose we devise a relation from Q to R that takes every integer q in Q and doubles it to get an even integer r in R. (No two girls have the same name. which is a proper subset of Q. The solutions in the appendix may not represent the only way a problem can be figured out. If we think of the mapping in Prob. Let A be the set of all members of Alpha at that time. This determines how the dance partners will be assigned. Imagine that you and I were once members of Internet Network Alpha. To generate a rational q from a nonzero integer z. from the set of girls to the set of boys) what type of mapping is it? 4. B. was well deserved. The maximal domain is Q. we were kicked out of Internet Network Alpha. we find the fractional equivalent of q in lowest terms. a punishment which. which is a proper subset of R. We were not totally honorable characters. and the range is Zeven. we realize today.) The boys pick slips out of the jar. If you think you can solve a particular problem in a quicker or better way than you see there. you and I. Suppose that we create a relation between the set Q of rational numbers and the set Z of integers. You may (and should) refer to the text as you solve these problems. the set of integers. Suppose that we create a relation between the set Z of integers and the set Q of rationals. Suppose that we had the latest programs to defeat antispam software in other people’s computers. What type of relation from Q to R have we devised? 6. and then chop off the denominator. by all means try it! 1. the set of all even integers. If we let q be the independent variable and z be the dependent variable. We conspired to send a mass e-mail message (also called spam) to every single one of the 175. The co-domain is R. with every other boy taking two slips instead of only one. one at a time and one for each girl. Five of the boys end up dancing with one girl. 2 in reverse (that is.000 members of Internet Network Beta. and then puts the slip into a jar. To generate an integer z from a rational q. what type of mapping is this? 2.Practice Exercises
221
Practice Exercises
This is an open-book quiz. so we succeeded in our dubious quest. and then puts the slip into a jar.

we say that q is not defined.222 Mappings. Also suppose that the domain of the new relation is the set of nonnegative reals. why not? Is the relation injective? Is it surjective? Is it bijective? 10. Consider the relation y = x 4. If we let z be the independent variable and q be the dependent variable. we say that q = 0 by default. Is this inverse relation a function? If so. but with one exception. we divide 1 by z). is this relation injective? Is it surjective? Is it bijective? 8. consider a relation between the set Z of integers and the set Q of rationals. why? If not. If z = 0. Relations. 9 are transposed while leaving their names the same. Imagine that the values of the independent and dependent variables in Prob. Is this relation a function? If so. If we let z be the independent variable and q be the dependent variable. 7. why? If not. Again. If z = 0. and the range of the new relation is the entire set of reals. why not? Is the relation injective? Is it surjective? Is it bijective?
. Imagine that this relation works in the same way as the relation in Prob. where the domain is the entire set of reals and the range is the set of nonnegative reals. and Functions
z (that is. is this relation injective? Surjective? Bijective? 9.

In most Cartesian graphs. In a more generalized system called rectangular coordinates or the rectangular coordinate plane. both axes are linear.
. the two axes do not have to be graduated in the same increments. II.
If a point lies exactly on one of the axes or at the origin. If we move 1/4 of an inch along an axis and the value changes by 1 unit. The first. This means that for any given axis. and fourth quadrants are sometimes labeled I. It’s best to choose the increments so a graph is easy to read. second. then it is not in any quadrant.Two Number Lines
225
6 4 Second quadrant 2 First quadrant
–6
–4
–2 –2 –4 –6
2
4
6
Third quadrant
Fourth quadrant
Figure 14-3 The Cartesian plane is divided into
quadrants. third. and both axes are graduated in increments of the same size. while the value on the other axis changes by 25 units for every 1/4 of an inch. The increments we select for each axis depend on what sort of relation or function we want to graph. III. then that fact is true everywhere along that axis and it is also true everywhere along the other axis. and IV respectively. The quadrants are sometimes labeled with Roman numerals. they’re located like this: • • • • Quadrant I is at the upper right Quadrant II is at the upper left Quadrant III is at the lower left Quadrant IV is at the lower right
Axis increments In a true Cartesian coordinate plane. The value on one axis might change by 1 unit for every 1/4 of an inch. the change in value is always directly proportional to the physical displacement.

14-1.
Add 2 to the input Figure 14-4 shows some points plotted in Cartesian coordinates for the following relation.226 The Cartesian Plane
Are you confused?
The coordinate planes in Figs.1 unit per division.” we can increase the numbers on one or both scales. draw a Cartesian plane on a piece of graph paper. it will move to the third. Then plot a specific point or two in each quadrant. If it starts out in the second quadrant. so the point does not fall on either axis.
Here’s a challenge!
Imagine an ordered pair (x. 13:
y=x+2 These points lie along a straight line. What will happen to the location of the point if you multiply both x and y by −1?
Solution
The point will move diagonally to the opposite quadrant. y). When we’ve plotted enough points so we’re reasonably sure we know what the graph will look like. as long as we draw the coordinate system so it’s easy to read. it will go “kitty-corner” across the coordinate plane.0002 unit per division! Our use of 6 increments on each of the four scales is arbitrary. If we want to show graphs “far out. it will move to the second. it will move to the first.0012 to 0. or from −3. we can go from −60 to 60 in increments of 10 units per division. You have plotted its point on the Cartesian plane. Neither x nor y is equal to 0.
If you have trouble envisioning this. In other words. and then plot the new points. Instead of going from −6 to 6 in increments of 1 unit per division. The line is the graph of the relation. Calculate how the x and y values change when you multiply both of them by −1. We can plot more points for the relation.0012 in increments of 0. and they will always lie along the same straight line.000 to 3. and plot the ordered pairs as points. or from −0. as follows: • • • • If it starts out in the first quadrant.000 in increments of 500 units per division.6 to 0.6 in increments of 0. and 14-3 show values only to up to ±6 for each variable.
Three Relations
To graph a relation in Cartesian coordinates. We might go from −0. it will move to the fourth. we can pick a few numerical values for the independent variable. This line or curve is the actual graph. If it starts out in the third quadrant.” we can make the numbers on the scales smaller.
. If we want to graph something “close in. 14-2. which was the first one we evaluated in Chap. calculate the resulting values for the dependent variable.
If it starts out in the fourth quadrant. we can connect the points with a smooth line or curve. We can have more or fewer.

” With simple relations such as those in this chapter.Three Relations y 6 4 2 (0. as follows:
y = ±(x1/2) Figure 14-5 shows some points for this relation. a few points are enough. which we’ll study in Part 3.
Positive/negative square root The second of the three relations we evaluated in Chap.–3) –4 –6 –4 –2 –2 2 4 6 (4.
Are you confused?
How many points must you plot before you truly know what the graph of a relation looks like? The best answer is. 14-5. but without the negative values of y. along with the curve that connects them. It is characteristic of quadratic equations. 13 took the positive or negative square root of x.6)
227
Figure 14-4 Cartesian graph of the relation y = x + 2. along with the curve that connects them. With more complicated relations. 14-6. It appears as the nonnegative half of the curve in Fig. “It depends.
Nonnegative square root The third relation we looked at in Chap. you might have to plot many points before the complete graph can be
. This curve is called a parabola.2) x –6 (–5. 13 was the same as the second one. It involved taking the nonnegative square root:
y = x1/2 Some points for this relation are plotted in Fig.

Now we’re operating in two dimensions instead of one. In the second case.” It works like the “number reflector” for generating negative numbers from positive numbers on the number line. the “point reflector” intersects every point-connecting line exactly in the middle. we say that the “point reflector” is a perpendicular bisector of any line connecting a point with its inverse point. There are computer programs that generate detailed graphs of relations by plotting millions of points and connecting them by means of a scheme called curve fitting. Technically. we must transpose the values of the variables without changing their names. the domain becomes the set of all reals. 3. as a “point reflector. let’s figure out the equations for these inverse relations. we’re ready to graph them. Figure 14-7 shows how this works. the domain is the set of nonnegative reals. Now that we know the equations for the inverse relations. The three original relations are y=x+2 y = ±(x1/2) y = x1/2 In Chap. In the first case. We can imagine this line.) For any point that’s part of the graph of the original relation. In the third case. we can find its counterpart in the graph of the inverse relation by going to the opposite side of the “point reflector. We must also transpose the domain and the range. and the range is the set of all reals. In the third case. we get x=y+2 x = ±(y1/2) x = y1/2 When we manipulate these equations to get x in terms of y. We draw the line representing all points where the independent and dependent variables have the same value. we were given the domains and ranges for these relations.
Solution
First. and the range becomes the set of nonnegative reals. The line connecting a point in the original graph and its “mate” in the inverse graph is perpendicular to the “point reflector. A simple trick makes it easy to graph the inverse of any relation.
. To do that. The plus sign in the last equation means that we consider it only for nonnegative values of x. the domain and range are both the set of nonnegative reals.” exactly the same distance away. the domain and range both remain the set of nonnegative reals. 13.
Here’s a challenge!
Draw graphs of the inverses of the three relations in this section. the domain and range are both the entire set of reals. they both remain the entire set of reals. because negative values of x aren’t part of the domain. the results are y=x−2 y = x2 y = (+x)2 When we transpose the domain and the range from the original relations in the first case.” In addition. (We devised that gimmick in Chap. Switching the values of the variables by reversing their positions in the original equation. In the second case. represented by the equation y = x in these examples.Three Relations
229
determined.

as well as the domain and range.–6) 2 (0. but the values of the variables.” That moves every point in the graph of the original relation to its new position in the graph of the inverse. we “flip the whole graph over” along a “hinge” corresponding to the “point reflector. 14-5. have been!
y 6 4 (5. These graphs show the inverses of the original three relations.
. 14-9. and 14-10 respectively. Note that the positions of the x and y axes have not been switched. and 14-6.
When we want to graph the inverse of a relation.3) 2 x –6 –2 –2 –4 (–4.14-8. When we do this to the graphs from Figs. we get the graphs in Figs. 14-4.–2) 4 6
Figure 14-8 Cartesian graph of the relation y = x − 2.230 The Cartesian Plane
y 6 4 Inverse Original 2 y=x Original "Point reflector" line Inverse
x –6 Inverse –4 –2 –2 Original –4 –6 Original Inverse 2 4 6
Figure 14-7 Any point in the graph of the inverse of a
relation can be located on the basis of its “mate” in the graph of the original relation.

and the curve connecting them is drawn.
Add 1 to the input Figure 14-11 shows some points plotted in Cartesian coordinates. Remember.
. for the following function:
y=x+1 These points lie along a straight line. 14-9. 14-12. a function is nothing more than a relation with special properties! In this section.
Square the input In Fig.
Cube the input Figure 14-13 shows what happens when the independent variable is cubed rather than squared. 13. The only difference here is that the line is exactly 1 unit lower on the coordinate plane.232 The Cartesian Plane
Three Functions
The process for graphing a function is the same as it is for graphing a relation. several points are plotted. for the function
w = v2 This graph is the same curve as the one shown in Fig. along with the curve connecting them. 14-4. Note the similarity between this graph and the one shown in Fig.1) x –6 –2 (–4. but the variable names are different.5) 4 2 (0.–3) –4 –6 2 4 6
Figure 14-11 Cartesian graph of the function y = x + 1. are plotted for the function
u = t3
y 6 (4. Several points. along with the straight line that connects them. we’ll look at the graphs of the three functions we evaluated in Chap.

If you think you can solve a particular problem in a quicker or better way than you see there. You’ll get a chance to do that in the last three Practice Exercises. merely by looking at their graphs. Draw a vertical line somewhere on the graph. where the vertical line doesn’t intersect the graph at all. both axes are linear. Sometimes—maybe all the time—this vertical line will intersect the graph of the relation. Imagine moving this horizontal line up and down.
Practice Exercises
This is an open-book quiz. 14-13. so the point does not fall on either axis. What happens to the location of the point if we multiply y by −1 and leave x the same?
. Draw a horizontal line parallel to the independent-variable axis. You’ll see that the line in Fig. or even large regions. you learned that a function never maps a single value of the independent variable to more than one value of the dependent variable. You can use this fact to determine whether or not a given relation is a function by looking at its graph. and suppose we have plotted its point on the Cartesian plane. 14-11 checks out. by all means try it! 1. You may (and should) refer to the text as you solve these problems. “Vertical” in this context means “parallel to the dependent-variable axis. B.234 The Cartesian Plane
Note the difference in “magnification” between the t and u axes. the movable vertical line must never intersect the graph at more than one point. so the point does not fall on either axis. Imagine an ordered pair (x.) This trick can be called the vertical-line test.) Now conduct this test on the graphs shown in Figs. This difference makes the graph fit nicely into the available space. Neither x nor y is equal to 0. 14-12. 14-11.
Solution
You can conduct a horizontal-line test on the graph of a relation to see if its inverse is a function.
Here’s a challenge!
How can you tell. and suppose we have plotted its point on the Cartesian plane. y). 14-12 fails the horizontal-line test! That means that the inverse of that relation is not a function. 13. Imagine an ordered pair (x. y). and 14-13. For the relation to qualify as a function. so the inverse of this relation is a function. Neither x nor y is equal to 0. this movable line must never intersect the graph of the original relation at more than one point. Even though an increment on the u axis represents 10 times the numerical change as an increment of the same length on the t axis.” Imagine moving this vertical line to the right and left. What happens to the location of the point if we multiply x by −1 and leave y the same? 2. (It’s okay if there are places or regions where the horizontal line doesn’t intersect the graph at all. The same is true for the relation graphed in Fig. For the inverse to qualify as a function. which of the three relations in this section have inverses that are functions? Don’t actually graph the inverses. But the curve in Fig. (It’s okay if there are places. The solutions in the appendix may not represent the only way a problem can be figured out.
Are you confused?
In Chap. Don’t hurry! You’ll find worked-out answers in App.

14-11. 6y)? Where. Does this equation represent a function of x ? 8. How can we use the horizontal-line test on a graph to determine whether or not a given numerical value is in the range of a function or relation? 6. Does this equation represent a function of x ? 7.
. and then plot the graph from scratch. Sketch a graph of the equation y = |x | for all real numbers x. 9. y /4)? 4. Imagine an ordered pair (x. 14-12. Derive the inverse using algebra. Where. y). will we find the point for (x /4. 14-13. Sketch a graph of the inverse of y = x + 1. y). y). in relation to the point for (x. in relation to the point for (x. Sketch a graph of the inverse of w = v 2. Don’t use the “point reflector” scheme from Fig. will we find the point for (6x. and suppose we have plotted its point on the Cartesian plane. Sketch a graph of the inverse of u = t 3. How can we use the same test on a graph to determine whether or not a given numerical value is in the domain? 5. Do this by applying the “point reflector” scheme to Fig.Practice Exercises
235
3. Do this by applying the “point reflector” scheme to Fig. Sketch a graph of the equation y = |x + 1| for all real numbers x. 10. The vertical-line test can be used to see whether or not a graph portrays a function.

y2) = (3. 4) and (x2.
y 6 (–2.Slope-Intercept Form
237
Sometimes the slope of a straight line is informally called rise over run.4) (3.5)
2 x –6 –4 –2 –2 –4 –6 m= y/ x = (5 – 4) / [3 – (–2)] = 1/5 2 4 6
Figure 15-1 The slope of a line can be calculated from
the coordinates of two points on that line.
Try two points! Suppose we see a line in the Cartesian plane. the dependent variable is on the vertical axis. 5) The slope is ( y2 − y1). which we call Δy. 15-1. y1) = (−2. and are able to locate two points on it and determine their exact coordinates:
(x1. and we move to the right. divided by (x2 − x1). This notion works as long as the independent variable is on the horizontal axis. which we call Δx : m = Δy /Δx = (y2 − y1)/(x2 − x1) = (5 − 4)/[3 − (−2)] = 1/(3 + 2) = 1/5 This situation is illustrated in Fig.
.

you learned the standard form for a first-degree equation in one variable. This method can be more convenient than rearranging everything into SI form or drawing a graph. That’s because both the numerator and the denominator end up being additive inverses (exact negatives) of what they were before. but we can’t reverse the internal sequence of either of the ordered pairs defining those points!
What is the intercept? When we talk about the SI form of a straight line in the Cartesian plane. If y is the dependent variable. we can figure the slope going from the first point to the second. That’s what is usually meant when we work with the SI form of an equation when graphing it in the xy-plane. If the variable is x. y1) and (x2. the term intercept refers to the value of a variable at the point where the line crosses the axis for that variable. 12. An intercept can be thought of as an ordered pair where one of the values (the one on the axis not being intercepted) is 0. and solve for the remaining variable to get its intercept. y2) = (−2. or going from the second point to the first. 5) and (x2. then we often talk about the y-intercept. 15-2. but all by itself it doesn’t give us any visual reinforcement of the situation. Putting it together In Chap. But we must be careful not to confuse the coordinates.238 Graphs of Linear Relations
Switching the order We can switch the points (x1. Calculating. we get m = Δy /Δx = (y2 − y1)/(x2 − x1) = (4 − 5)/(−2 − 3) = −1/(−5) = 1/5 When we know the coordinates of two points on a line. We can reverse the external sequence in which we work with the points. 4) The slope is again equal to Δy /Δx. the standard form is
ax + b = 0
. Let’s take
(x1. y2) and still get the same slope when we calculate it as above. it doesn’t matter. Two examples are shown in Fig. y1) = (3. We can plug 0 into a linear equation for one of the variables.

. you get an equation for a linear function where y is the dependent variable and x is the independent variable: y = ax + b As things work out. and the relation is not a function. If you substitute y for 0 and then transpose the left and right sides.Slope-Intercept Form y 6 4 2 y-intercept is 3 x –6 –4 –2 –2 –4 –6 2 4 6 Slope is positive
239
y-intercept is –2
Slope is negative
Figure 15-2 Two examples of y-intercept points for
straight lines. the line that ramps downward as we move to the right has negative slope. you can write y = mx + b This is the classical expression of the SI form for a linear function. The graph of a linear function must be a nonvertical line. and the constant b is the y-intercept. they are parallel to the dependent-variable axis. Note that all the lines are vertical. Figure 15-3 shows some examples.
where a and b are constants. otherwise it would fail the vertical-line test in the worst possible way! Whenever you see a linear relation that simply says x is equal to some constant. the constant a is the slope of the graph. then the slope is undefined. The line that ramps upward as we move to the right has positive slope. then you know that relation is not a function of x.
Are you confused?
If the graph of a linear relation is a vertical line. Because the slope is usually symbolized by m instead of a.

. Subtracting 8x from both sides gives us 4y = −8x + 12 Dividing each side by 4 puts it into SI form: y = (−8x)/4 + 12/4 = −2x + 3 The slope is −2. To plot the line. 3). and graph it on that basis: 8x + 4y = 12
Solution
We must rearrange this equation to get y all by itself on the left side of the equality symbol.240 Graphs of Linear Relations y 6 4 2 x –6 –4 –2 –2 –4 –6 x = –5 x = 3/2 x = 5. and the y-intercept is 3. That gives us a point with coordinates (0. because they are all straight lines parallel to the y axis. We plot the y-intercept point on the y axis at the mark for 3 units. Figure 15-4 shows the graphing process.78 2 4 6
Figure 15-3 These linear relations are not functions of x. and an expression containing only x and one or more constants on the right side.
Here’s a challenge!
Put the following equation into SI form as a linear function of x.
The slopes of the graphs are undefined.

there’s a standard form for a two-variable linear equation.
we must know the coordinates of another point.
.Slope-Intercept Form y Slope = –2 6 Move to right by 3 units y-intercept is 3 2 Move up by –6 units x –6 –4 –2 –2 –4 –6 6 4 which is down by 6 units
241
Figure 15-4 Graph of the equation 8x + 4y = 12. We plot this point. We have now found a second point on the graph. which is equivalent to moving straight down by 6 units. Our new point is 3 units to the right and 6 units below the y-intercept point. and a. and then we connect the two points with a solid line to obtain the graph. We can find one by moving horizontally to the right by any number of units we want (call it n units). where m is the slope. We should move far enough to the right so the two points will be well separated. and then moving straight up from there by mn units. so its coordinates are (3. We extend the line somewhat beyond the points in either direction. That gives us the point (3. −3). b. keeping in mind that the true. 3) by 3 units. This
graph can be drawn easily when we morph the equation into its SI form y = −2x + 3. complete line (in the mathematical cosmos) extends infinitely far in each direction!
Here’s another challenge!
Just as there is a standard form for a first-degree equation in one variable. expressing y as a function of x (as long as b. shown as an open circle to indicate that it’s not actually part of the graph. Show how this equation can be rearranged into the SI form. Let’s move to the right from (0. y is the dependent variable. Here it is: ax + by + c = 0 where x is the independent variable. That will make it easy to draw the graph accurately. Then we move straight up by mn = −2 × 3 = −6 units. is not equal to 0). the coefficient of y. and c are constants. 3).

2). m is the slope. This only works if the constant b (the coefficient of y) is not equal to 0. y0) are the coordinates of a known point on the graph. and the y-intercept is equal to −c /b. that’s the important thing! The coefficient of y cannot equal 0 in the original equation. Our task is to draw a graph of the function. Conversion of a general two variable linear equation to SI form. Later in this chapter. we’ll figure out how this form is derived:
y − y0 = m(x − x0) where x is the independent variable. Let’s call this the PS form. Let’s begin by assigning x0 = −1 and y0 = 2. and if we also know the slope of the line. Here. In this result. y is the dependent variable. As the name suggests. When we plug these numbers into the stan-
. The independent variable is x. the familiar m for slope is replaced by −a /b. and the familiar b for slope is replaced by −c /b.
An example Suppose we’re told that there’s a linear function whose graph contains the point (−1. The graph in such a case would exist. and the dependent variable is y.
The form Here is the standard PS form for a linear function. We are also told that the slope of the graph is 2. putting the equation into strict SI form
Solution
Table 15-1 is an S/R derivation of a SI equation from the standard form of a two-variable linear equation.
Statements ax + by + c = 0 ax + by = −c by = −ax − c Require that b ≠ 0 y = (−ax − c)/b y = −ax /b − c /b y = (−a /b)x − c /b y = (−a /b)x + (−c /b) Reasons This is the equation we are given Subtract c from each side Subtract ax from each side We’re about to divide through by b Divide through by b Right-hand distributive law for division over subtraction Rearrange to define the coefficient for x Change subtraction to negative addition. that would cause both the slope and the y-intercept to be undefined. and (x0. we can draw the graph of a function if we know the coordinates of any single point on the line. the slope is equal to −a /b. The result is in the correct form.
Point-Slope Form
Another common way to express a linear function is known as the point-slope form. so it would not represent a function of x. but it would be a vertical line.242 Graphs of Linear Relations
Table 15-1.

When we plot these two points on the plane and draw a straight line through them both. and the slope is equal to 2.2) (1. we get the graph shown in Fig.
. we get y − 2 = 2[x − (−1)] which simplifies to y − 2 = 2(x + 1) We can now find another point on the graph by plugging in a value for x and solving for y. We solve for y in steps: y − 2 = 2(1 + 1) y−2=2×2 y−2=4 y=6 This tells us that (1.Point-Slope Form
243
dard PS equation. We already know that (−1. 2) is on it. 15-5.
y 6 4 (x0. We only need one more point to determine the straight line in the Cartesian plane that represents this function.y0) = (–1. Let’s try x = 1. 6) represents a point on the graph.6)
x –6 –4 –2 –2 m=2 –4 –6 2 4 6
Figure 15-5 Graph of a linear function based on the
knowledge that the point (−1. 2) is on the line.

244 Graphs of Linear Relations
Table 15-2 Conversion of a linear equation from PS to SI form. 15-6.
. but we do know the coordinates of some point in one of the quadrants.
Here’s a challenge!
Look again at the general PS form for a linear equation.
Solution
Table 15-2 is an S/R derivation that shows how this can be done. x0. The y-intercept. The PS form is handy when we don’t know the y-intercept of a graph. whose graph has a known point with coordinates (x0y0) and a slope m. b) on the line. If we move away from (0. Then we’ll derive a standard form for a linear equation based on two known points. or Δy /Δx. as shown in Fig. we can easily write down an equation representing the function using the PS form.
Statements y − y0 = m(x − x0) y − y0 = mx − mx0 y = mx − mx0 + y0 y = mx + (−mx0) + y0 y = mx + (−mx0 + y0) y = mx + (y0 − mx0) Reasons This is the equation we are given Distributive law of multiplication over subtraction Add y0 to each side Change subtraction to negative addition Grouping of addends Simplify second addend on right side
Are you confused?
The PS form of a linear function is actually a generalized version of the SI form. If we are told only those coordinates and the slope. and y0 are constants. Remember that m. We can then draw the graph by finding another point using that equation. the slope is always equal to the difference in the y value divided by the difference in the x value. b). called b in the classical expression of the SI form. and connecting the two points with a straight line.
Known slope and y -intercept Imagine a line in Cartesian coordinates that has slope m and crosses the y axis at the point (0. turns out to be the quantity ( y0 − mx0).
Equations from Graphs
Let’s derive the SI and PS forms of linear equations by looking at how their graphs behave generally. and where x is the independent variable and y is the dependent variable: y − y0 = m(x − x0) Convert this equation into SI form.

b) to some point (x. y) will be 0 + Δx. because we have moved Δx units horizontally from a point where x = 0. then we will have demonstrated how y is a function of x.b) m= y x x +x
–y
Figure 15-6 The SI form of a linear equation can be derived
from this generic graph. y) will be b + Δy. y) on the line by going Δx units to the right and Δy units upward.
Suppose we move from (0. We can express Δy in terms of the slope m and the increment Δx by morphing the formula that defines slope. getting y = b + mΔx
. is m = Δy /Δx Multiplying through by Δx. If we can get an equation that allows us to calculate y in terms of x for the arbitrary point (x.Equations from Graphs +y x=0+ x y=b+ y (x. The x coordinate of the point (x. The y coordinate of the point (x. because we have moved Δy units vertically from a point where y = b. we get m Δ x = Δy Now remember that y = b + Δy We can substitute mΔx for Δy in this equation. y).y)
245
y –x (0. we’ll also get the SI form of the equation for the line. once again. As things turn out. That formula.

The y coordinate of the point (x. how slope is defined:
m = Δy /Δx
+y x = x0 + y = y0 + x y (x. we have moved from the y axis (where x = 0) horizontally by x units. If we move away from (x0. y) will be x0 + Δx. just as we did when we derived the SI equation. we can reverse the order of the addends to state it as y = mx + b
Known point and slope Imagine a line in the Cartesian plane that passes through a point whose coordinates are (x0y0).246 Graphs of Linear Relations
But in this situation. y0) to some arbitrary point (x. Suppose the line has slope m as shown in Fig. Because of this lucky coincidence. 15-7. y) will be y0 + Δy. we can substitute x for Δx in the above equation. Now remember.
. by traversing the increment Δx. y) on the line. because we have moved Δx units horizontally from a point where x = x0. y0 ) x +x
m=
y
x
–y
Figure 15-7 The PS form of a linear equation can be derived
from this generic graph. where x0 and y0 are known constants.y)
y –x (x0 . once again. Let’s go from (x0. the slope is always equal to Δy /Δx. Δx is exactly equal to x ! That’s because. getting y = b + mx If we want to be picayune. because we have moved Δy units vertically from a point where y = y0. The x coordinate of (x. y0) along the line.

” Think of what happens when we move to the right in the Cartesian plane. for example. That gives us y = y0 + mΔx Now we can see from Fig. “Nothing special. is y = mx + b This equation has a plus sign whether b is positive or negative. once again. we have mΔx = Δy Observe that in Fig. The standard form. 15-7 that x = x0 + Δx Subtracting x0 from each side. then Δy is always negative as we move to the right.” Similar sign-related confusion can occur when we work with the SI form of a linear equation. In this context. If m = 3 and b = −2. If we keep adding a positive Δy. we go lower and lower. Either way. then Δy is always positive as we move to the right. “What happens when the slope of the line is negative?” The answer is. If we keep adding a negative Δy. we add Δy when we move to the right. There’s good reason to wonder. If the slope is negative. y = 3x + (−2)
. we get y = y0 + m(x − x0) Subtracting y0 from each side gets us to the PS form y − y0 = m(x − x0)
Are you confused?
The preceding two examples show lines with positive slope. If the slope of a line is positive. “higher” means “in the positive y direction. y = y0 + Δy Let’s substitute mΔx for Δy here. we obtain x − x0 = Δx Substituting (x − x0) for Δx in the equation for y in terms of y0 and mΔx. as long as we’re careful. 15-7. we go higher and higher.Equations from Graphs
247
As before.” and “lower” means “in the negative y direction.

x2. if m = 5. y1) and (x2. for example.248 Graphs of Linear Relations
which is the same as y = 3x − 2 When we work with the PS form. and y0 = −8. we come across another “sign-rigid” situation. Then Δy = y2 − y1 and Δx = x2 − x1 The slope is m = Δy /Δx = (y2 − y1)/(x2 − x1)
. y1. It is Δy /Δx. y0) = (−5.
Here’s a challenge!
Imagine a straight line that passes through two points whose Cartesian coordinates are (x1. and pays no heed to whether y0 or x0 happen to be positive or negative. The line has a negative slope. Derive an equation for this line in terms of the independent variable x and the dependent variable y. Let’s move to the right. but that doesn’t make any difference in the way things will turn out. y2). and y2 to be constants.
Solution
Figure 15-8 shows a generic example of this situation. y = 2x − 4 then the y-intercept is b = −4. from the point (x1. y2). Let’s start by calculating the slope of the line. x0 = −4. but with minus signs instead of a plus sign! The general form of the equation is always y − y0 = m(x − x0) It contains two minus signs. For example. Consider x1. 3). If we see y − 3 = −4(x + 5) then the graph contains a point whose coordinates are (x0. y1) to the point (x2. then y − (−8) = 5[x − (−4)] which is the same as y + 8 = 5(x + 4) We must always pay close attention to signs when working with the standard forms of linear equations. Call this the two-point form of a linear equation. It’s easy to get them wrong! If we see.

We can substitute x1 for x0. When the coordinates for the points are given as numbers. we can plug them in and get the equation by means of straightforward arithmetic. and y1 for y0 in the classical PS equation to get y − y1 = m(x − x1) Substituting (y2 − y1)/(x2 − x1) for m.
Now let’s use the PS form to derive an equation for the line. Don’t hurry! You’ll find worked-out answers in App.
Practice Exercises
This is an open-book quiz. The solutions in the appendix may not represent the only way a problem can be figured out.Practice Exercises +y (x1 . B. y2 ) +x x y
249
m=
y
x
–y
Figure 15-8 A two-point form of a linear equation can be
derived from this generic graph. we obtain y − y1 = [(y2 − y1)/(x2 − x1)] (x − x1) which can be rearranged to y − y1 = (x − x1)(y2 − y1)/(x2 − x1) This is a mess. y1 ) x2 = x 1 + y2 = y 1 + y –x x (x 2 . Let’s use (x1. but it’s the best we can do when we aren’t given the slope directly. If you think you can solve a particular problem in a quicker or better way than you see there. The good news is that most of the values in this equation are constants. by all means try it!
. Either point will work. You may (and should) refer to the text as you solve these problems. We have two points to choose from. y1).

” How can she say this without drawing the graphs? Under what circumstances will she be right? Under what circumstances will she be wrong? 7. −6). 2. −12). “The two lines we talked about will intersect somewhere on the t axis. 1 through 4 using the simplest possible method. Calculate the slope of the line in Prob. showing that the slope doesn’t depend on which way we move along the line.250 Graphs of Linear Relations
1. 1. Use the results of the last challenge. and label the point of intersection. y) coordinates that passes through the two points (2.
.” She’s right! What is the exact point of intersection? 8. Point P is defined by (−1. Our advisor. −10) and (6. 8) and (0. 10. 6. Find an equation for the line in Cartesian (x. Sketch a graph of the linear equation discussed in Probs. 4. 2). “When graphed. −4). who introduced herself in Prob. where the independent variable is u and the dependent variable is v. 9. 6. Put the equation into PS form. and 3 in SI form. these equations will produce lines oriented at a 90° angle with respect to each other. Derive an equation of the line described in Probs. 1 on the basis of going in the direction from Q to P. Use the results of the last challenge. and indicate their values. What is the slope of the line connecting these two points if we go in the direction from P to Q ? 2. Graph the two lines we discussed in Probs. goes on to make the claim. Suppose we see two equations where s is the independent variable and t is the dependent variable: t=s+5 and t=5−s Someone says. 1 and 2 in PS form. Imagine two points P and Q plotted on the Cartesian plane. and Q is defined (2. Put the equation into SI form. Derive an equation of the line described in Probs. 6 and 7. 5. Label the slope as m and the v-intercept as b. y) coordinates that passes through the two points (−6. 3. Find an equation for the line in Cartesian (x.

we get −9x + 4 = −41 We can subtract 4 from each side to obtain −9x = −45 Finally. we get −y = −7x + 41 Multiplying through by −1 gives us y = 7x − 41
Mix the right sides and solve Now let’s put the right side of the first SI equation on the left side of an equals sign.−6) if we imagine x as the independent variable and y as the dependent variable. When we subtract 7x from each side. That gives us
y = −2 × 5 + 4 = −10 + 4 = −6 We have used algebra to find that x = 5 and y = −6. Let’s use the first one. we divide through by −9 to get x=5
Substitute and solve again To solve for y.252 Two-by-Two Linear Systems
Dividing through by 4. we can take either of the SI equations and plug in 5 for x. and the right side of the second SI equation on the right side of the same equals sign. We can express this as the ordered pair (5. This produces a first-degree equation in one variable:
−2x + 4 = 7x − 41 Now let’s solve this for x. we get y = −2x + 4 Now for the second equation. We begin with 7x − y = 41 Subtracting 7x from each side.
.

Are you confused?
Once you’ve solved a two-by-two linear system (or think you have). if we see the equation
x − y = 10 then we can say that both of the following are SI equivalents of it: y = x − 10 and x = y + 10 In the first case. too! You can now be confident that the solutions are right. What is the airspeed of the plane? What is the speed of the wind relative to the earth?
. In the second case. The solution to the system originally stated on page 251 appears to be x = 5 and y = −6. When flying right along with that same wind at the same airspeed (speed measured with respect to the surrounding air). Is it. But a true SI equation always has the dependent variable alone on the left side of the equals sign. really? Check the first original equation: 8x + 4y = 16 8 × 5 + 4 × (−6) = 16 40 − 24 = 16 16 = 16 That checks out fine! Now for the second original equation: 7x − y = 41 7 × 5 − (−6) = 41 35 + 6 = 41 41 = 41 It works out here. we treat y as the dependent variable. you should consider your solutions tentative until you’ve plugged them into both of the original equations and worked out the arithmetic to be sure that they’re correct. we treat x as the dependent variable. and the independent variable on the right side along with constants that represent characteristics of a graph.
Here’s a challenge!
While flying directly into a high-altitude wind. the plane has a groundspeed of 990 km/h. For example. an airplane has a groundspeed (speed measured with respect to the earth) of 750 kilometers per hour (km/h).Morph and Mix
253
Two versions of the SI form In a two-by-two system. it often doesn’t matter which variable we consider independent and which one we consider dependent.

Therefore x − y = 750 Let’s put this equation into SI form. and let y represent the speed of the wind. we obtain x = −x + 1. getting y = −x + 990 Now let’s mix the right sides of these two SI equations: x − 750 = −x + 990 When we add 750 to each side. getting y = x − 750 = 870 − 750 = 120 This tells us that the wind is blowing at 120 km/h with respect to the earth.254 Two-by-Two Linear Systems
Solution
Let x represent the airspeed of the plane. both in kilometers per hour.
. we can subtract x from each side. We can use the first one. That means x + y = 990 To morph this into SI form. When the plane flies against the wind.740 We can add x to each side to get 2x = 1.740 Finally we divide through by 2. the wind adds to its groundspeed. Subtracting x from each side gives us −y = −x + 750 Multiplying through by −1 tells us that y = x − 750 When the plane flies with the wind. Let’s plug this value into one of the SI equations. discovering that x = 870 Now we know that the airspeed of the plane is 870 km/h. the wind takes away from its groundspeed.

That allows us to solve for the other variable. It’s also called the addition method. getting
4t + 10u = −14
. Let’s use the form of the first equation. getting −4t + u = −3
Eliminate the first variable Our first objective is to make t vanish when we add multiples of the equations. The second equation can be morphed into that form by subtracting 4t from each side. Here we go: x + y = 990 870 + 120 = 990 990 = 990 and x − y = 750 870 − 120 = 750 750 = 750 We can now state these solutions with confidence. one of the variables disappears. We can multiply the first original equation through by 2.Double Elimination
255
These values should be checked by plugging them into both of the original equations to be sure they’re correct. at least until the wind changes or the airplane alters its cruising speed!
Double Elimination
There’s another way to solve two-by-two systems of linear equations. we must get both equations in the same form. We can morph one or both equations so that when we add them in their entirety. and solve for the one!
Get in the same form Let’s solve the following two-by-two linear system using double elimination. We can then do the same thing to make the other variable disappear. first for u (by eliminating t) and then for t (by eliminating u):
2t + 5u = −7 and u = 4t − 3 Before we go any farther. which I like to call double elimination.

Eliminate the second variable Now let’s look at the first original equation and the second morphed equation again.u) = (4/11. Let’s multiply the second equation through by −5. we see that u = −17/11.
Are you confused?
Let’s plug these results into the original equations to be sure they’re accurate. The first equation figures out this way: 2t + 5u = −7 2 × 4/11 + 5 × (−17/11) = −7 8/11 + (−85/11) = −7 −77/11 = −7 −7 = −7
. That morphs it into 20t − 5u = 15 We can add the two equations in their entirety: 2t + 5u = −7 20t − 5u = 15 22t = 8 This is easily solved to get t = 8/22.256 Two-by-Two Linear Systems
Adding the two morphed equations in their entirety causes the variable t to vanish: −4t + u = −3 4t + 10u = −14 11u = −17 When we divide this result through by 11. which reduces to 4/11. We can now state the solution to this system as an ordered pair (t. They are
2t + 5u = −7 and −4t + u = −3 Our goal this time is to find a way to make u vanish when we add multiples of the equations.−17/11).

assuming that (db − ae) ≠ 0: y = (dc − af )/(db − ae) Now let’s cause y to disappear so we can solve for x. and multiply the second equation through by −b. we can invoke the distributive law “backward” in the left side of this result to get (db − ae)y = dc − af We can solve for y if we divide through by (db − ae). We can be confident that we’ve found the correct solution to the original twoby-two linear system. they vanish.
Here’s a challenge!
Derive a general formula using double elimination that solves the following two-by-two linear system for the variables x and y in terms of the constants a through f. We multiply the first original equation through by e. we must get coefficients for x that have the same absolute value but opposite sign in the two equations.Double Elimination
That works! The second original equation comes out like this: u = 4t − 3 −17/11 = 4 × 4/11 − 3 −17/11 = 16/11 − 3 −17/11 = 16/11 − 33/11 −17/11 = (16 − 33)/11 −17/11 = −17/11
257
That checks out as well. and multiply the second equation through by −a. Let’s multiply the first equation through by d. Here are the equations: ax + by = c and dx + ey = f
Solution
First. so when we add them. When we add the resulting equations in their entirety. let’s cause x to disappear so we can solve for y. we get dax + dby = dc −adx − aey = −af dby − aey = dc − af The terms dax and −adx are additive inverses. we get eax + eby = ec −bdx − bey = −bf eax − bdx = ec − bf
. To do this. Now. When we add the resulting equations in their entirety.

Consider this pair of equations:
−7v + w + 10 = 0 and 4v + 8w = −40 We can take the first equation and add 7v to each side. and then plug the number into a strategic spot to solve for the other variable. we begin by morphing one of the equations into SI form.
Rename one variable When we want to solve a two-by-two linear system by rename-and-replace. Next. This process is usually called the substitution method. we can apply the distributive law “backward” in the left side. Can you see why? What do you think will happen if a linear system has coefficients in the above form such that ae = bd ? You’ll get a chance to explore a situation like that in exercises 5. obtaining (ea − bd )x = ec − bf We can solve for x if we divide through by (ea − bd ). I like to call it rename and replace. assuming that (ea − bd ) ≠ 0: x = (ec − bf )/(ea − bd )
Are you astute?
Have you noticed that the “taboos” in the above derivations are actually two different ways of saying the same thing? They’re stated like this: (db − ae) ≠ 0
and
(ea − bd ) ≠ 0 Both of these inequalities are equivalent to the statement ae ≠ bd. 6. and 7 at the end of this chapter!
Rename and Replace
A two-by-two linear system can be unraveled by renaming one variable in terms of the other. getting w + 10 = 7v
. first-degree equation from the result. and then creating a single-variable.258 Two-by-Two Linear Systems
The terms eby and −bey are additive inverses. so they disappear from the sum. We solve that equation.

we get w = 7 × 2/3 − 10 Taking the product on the right side of the equals sign. The simplest approach is to use is the SI equation we derived in the first step:
w = 7v − 10 When we replace v by 2/3 here.
. which in this case is the second original. That gives us
4v + 8(7v − 10) = −40 The distributive law of multiplication over subtraction can be applied to the second addend on the left side of the equals sign to get 4v + 56v − 80 = −40 Summing the first two addends in the left side of this equation gives us 60v − 80 = −40 Adding 80 to each side. and changing 10 into 30/3 to obtain a common denominator.
Plug the number into the best place Now that we have solved for one of the variables. we obtain 60v = 40 This tells us that v = 40/60 = 2/3. we come up with w = 14/3 − 30/3 Now it’s a matter of mere arithmetic: w = (14 − 30)/3 = −16/3 We’ve derived the solution to this system: v = 2/3 and w = −16/3.Rename and Replace
259
Then we can subtract 10 from each side to obtain this SI equation with w playing the role of the dependent variable: w = 7v − 10
Make a first-degree equation The second step involves substituting our “new name” for w into the equation we haven’t touched yet. we can replace the resolved unknown with its solution in any relevant equation containing both variables.

Solution
Let’s tackle the second equation and get it into a form that expresses y in terms of x. First. getting 2y = 8x + 4 When we divide through by 2. The signs will be tricky. Here’s the first check: −7v + w + 10 = 0 −7 × (2/3) + (−16/3) + 10 = 0 −14/3 − 16/3 + 10 = 0 −30/3 + 10 = 0 −10 + 10 = 0 0=0 All right! Here’s the second check: 4v + 8w = −40 4 × 2/3 + 8 × (−16/3) = −40 8/3 − 128/3 = −40 (8 − 128)/3 = −40 −120/3 = −40 −40 = −40 All right again! Our solutions are correct. it won’t be too bad. we had better check our work to be sure the solutions we obtained satisfy both of the original equations. we get y = 4x + 2
.
Here’s a challenge!
Solve the following pair of equations as a two-by-two linear system using the substitution method: 3x − πy = −1 and −8x + 2y = 4 This process is going to be messy! If we remember that π is a plain old real number. though.260 Two-by-Two Linear Systems
Are you confused?
As always. we can add 8x to each side.

Rename and Replace
Now we can substitute (4x + 2) for y in the first original equation. all right! But we’ve found a real number that’s equal to x. But we must take a step back. we get 3x − 4πx = −1 + 2π We can use the distributive law “backward” to morph the left side of this equation. divided by that denominator. The idea is to get a common denominator. and get a simpler expression as a result.”) Our solution for y can now be rewritten as y = (−4 + 8π)/(3 − 4π) + 2(3 − 4π)/(3 − 4π) This gives us a sum of two fractions with the common denominator (3 − 4π). add some fractions. obtaining (3 − 4π)x = −1 + 2π Now we can divide through by (3 − 4π) to get x = (−1 + 2π)/(3 − 4π)
261
It’s a mess. We can plug this number into the SI equation we derived earlier. In mathematical terms. getting y = 4[(−1 + 2π)/(3 − 4π)] + 2 = (−4 + 8π)/(3 − 4π) + 2 Believe it or not. in advance. obtaining 3x − π(4x + 2) = −1 When we apply the distributive law on the left side of the equals sign. how a scheme like this will work. and then we can take two steps forward. we get 3x − (4πx + 2π) = −1 This is the equivalent of 3x + [−1(4πx + 2π)] = −1 which simplifies to 3x − 4πx − 2π = −1 When we add 2π to each side. 2 = 2(3 − 4π)/(3 − 4π) It takes some intuition to see. (With practice. this can be simplified. Let’s “complexify” the number 2 and write it as twice the denominator in the fraction above. you’ll develop this “sixth sense. Therefore: y = [(−4 + 8π) + 2(3 − 4π)]/(3 − 4π)
.

The ball strikes the pavement at 135 miles per hour (mi/h). If you think you can solve a particular problem in a quicker or better way than you see there. 3. You throw a baseball straight out in front of the car. One of them is 6 times the other.262 Two-by-Two Linear Systems
Applying the distributive law. The sum of two numbers is 100. moving opposite to the direction of the car. Then you throw another baseball directly backward. it’s also moving backward relative to the pavement!) How fast am I driving? How fast do you hurl the baseballs relative to the car? Forget about the possible effects of wind and gravity.
Practice Exercises
This is an open-book quiz. What are the two numbers? Use double elimination. The solutions in the appendix may not represent the only way a problem can be figured out. 4. at 15 mi/h. You’re on your own! Here’s a hint: (3 − 4π) divided by itself is equal to 1. The sum of two numbers is −83. we get our reward: y = 2/(3 − 4π) We’ve arrived at our solutions! They are: x = (−1 + 2π)/(3 − 4π) and y = 2/(3 − 4π)
How about some extra credit?
Plug the above numbers into the original equations for x and y. The second ball hits the highway. Their difference is 10. (The ball is not only moving backward relative to the car. Here’s a hint: Feel free to draw diagrams. we get y = (−4 + 8π + 6 − 8π)/(3 − 4π) When we add up the terms in the numerator here. exactly as hard as the first one. Don’t hurry! You’ll find worked-out answers in App. by all means try it! 1. The sum of two numbers is 44. What are the two numbers? Use the morph-and-mix method. and verify that the answers we got are correct. There are no other vehicles or living things in sight. 2. I’m the driver. B.
. You may (and should) refer to the text as you solve these problems. Imagine that you and I are traveling in a car on a level highway at constant speed. Their difference is 13. What are the two numbers? Use the morph-and-mix method.

1 using the rename-and-replace (substitution) method.Practice Exercises
263
5. Solve the two-by-two linear system given in Prob. Solve the two-by-two linear system given in Prob. Here are the equations: 2x + y = 3 and 6x + 3y = 12 Is something wrong with one or both of these equations? Why can’t we solve this system? 6. 10. You’ll get a meaningless result. Try it using the double-elimination method and see what happens. Look again at the general derivation for solving a two-by-two system by double elimination. 5 into SI form. 9. Attempt to solve the following pair of equations as a two-by-two linear system by substitution. Does this have anything to do with why the system described in Probs. 8. What does this tell us about their graphs? Would drawing the graphs provide any clues about why the system can’t be solved? 7. Remember the “taboo” concerning the constants. 5 and 6 can’t be solved? Plug in the numbers and see what happens. 4 using the rename-and-replace method. Put the two original equations from Prob. When we attempt to solve the following two-by-two linear system. we will fail. Why? s = 2r − 3 and −10r + 5s + 15 = 0
.

−41). 17-1. because the increments must be large (in this case 10 units per division). two of the points we found are close to the origin.4) x Each axis increment is 10 units Solution = (5. 4) and (5. Those SI equations. Figure 17-1 shows the graph of this system on the Cartesian plane. It’s difficult to plot points accurately when they’re so diverse.–56)
Figure 17-1 Graphs of 8x + 4y = 16 and 7x − y = 41
as a two-by-two linear system where the independent variable is x and the dependent variable is y. The ordered pairs for those points are (0. We Mixed. −41) and (5. We Can Graph
265
When we morphed the above two equations before mixing them on our way to a solution.
Are you confused?
In Fig.We Morphed. while the third point is far away. It’s also difficult to draw a line based on two points that are close together. once again. The lines intersect at the solution point where x = 5 and y = −6. −6). we put them into SI form.–41)
(30. If we want to
y 8x + 4y = 16 7x – y = 41 (0.
Connect the points We have determined that one line passes through the points (0.4) and (0. each increment represents 10 units.
. −6).–6)
(0. while the other line passes through the points (0. are y = −2x + 4 and y = 7x − 41 The y-intercepts are at 4 and −41. corresponding to the ordered pair (5. On both axes.−6).

the plane has a groundspeed of 990 km/h. we can plug in x = 30 and easily solve for y using the SI form: y = −2x + 4 = −2 × 30 + 4 = −60 + 4 = −56 That tells us that the point (30. The solution we obtained in Chap.266 Two-by-Two Linear Graphs
precisely draw the line through two points that are close together. and the wind is blowing at 120 km/h. When flying straight downwind at the same airspeed.
. 16 was x = 870 and y = 120. −750) and (0. 4) and (5. 990) on a Cartesian plane where x is the independent variable and y is the dependent variable. This point is far enough away from (0. giving us another point with the ordered pair (870. It is shown as an open circle in Fig. While flying straight into the wind. showing that the airspeed of the plane is 870 km/h. both in kilometers per hour. The reference points we found are well separated. an airplane has a groundspeed of 750 km/h. −6). For example. That’s how this two-by-two linear system looks when graphed. 16. −56) is on the line. in the equation 8x + 4y = 16 for the line through (0. and let y represent the speed of the wind. 4) so we can easily draw the line. They are y = x − 750 and y = −x + 990 The y-intercepts are −750 and 990. we can find a third point farther out on that line and use it. Let x represent the airspeed of the plane.
Here’s a challenge!
Revisit the airplane challenge from Chap. so we can plot the points (0.
Solution
Let’s write down the SI forms of the equations that we derived before we mixed them. 17-1.120). so the lines are easy to draw. Then x − y = 750 and x + y = 990 Graph these two equations. Figure 17-2 shows these three points and the lines through them.

–750)
x – y = 750
Each axis increment is 200 units
Figure 17-2 Graphs of x − y = 750 and x + y = 990
as a two-by-two linear system where the independent variable is x and the dependent variable is y. so we know that (0. The second equation is already in SI form. calling t the independent variable and u the dependent variable. On both axes. We Eliminated. We start with
2t + 5u = −7
. 16. Let’s graph this situation.
Find two points for each line As before. −3) is on the line. We have to work on the first equation a little. We Can Graph y x + y = 990 Solution = (870.We Added.990)
267
x
(0. using these points as the basis for drawing our lines.120) (0.
We Added. We Can Graph
In Chap. we solved the following two-by-two linear system for t and u using double elimination: 2t + 5u = −7 and u = 4t − 3 We found that t = 4/11 and u = −17/11. each increment represents 200 units. The u-intercept for its graph is −3. We Eliminated. let’s find the u-intercepts and the solution.

268 Two-by-Two Linear Graphs
When we subtract 2t from each side.–17/11)
(0. Our first line passes through (0. because it’s customary to list the independent variable first and the dependent variable after it. In Chap. 16. On both axes. −17/11).u).–7/5) Solution = (4/11. as can be seen by looking at the graph of this system (Fig.–13/5)
Figure 17-3 Graphs of 2t + 5u = −7 and u = 4t − 3
as a two-by-two linear system where the independent variable is t and the dependent variable is u. −7/5) and (4/11. −3) and (4/11. 17-3). The second line goes through (0. Note that the ordered pairs here are always of the form (t. two of these points are so close together that it’s hard to draw the line through them precisely. we get 5u = −2t − 7 We can divide through by 5 to obtain u = (−2/5)t − 7/5 The u-intercept for this line is −7/5.
. the point (4/11.–3)
(3. −17/11) is where the lines intersect.1) t 2t + 5u = –7 (0. −17/11). each increment represents 1/2 unit.
Connect the points We have the points we need to graph the lines. −7/5) is on it.
u
Each axis increment is 1/2 unit
u = 4t – 3
(1. Therefore. we found that the solution for the linear system was t = 4/11 and u = −17/11. Unfortunately. so the point (0.

The calculations are usually simple if we use integers for the “plug-ins. Finally. you can figure out the reasoning behind each step. we’ll draw the lines through the points. we get u = (−2/5)t − 7/5 = (−2/5) × 3 − 7/5 = −6/5 − 7/5 = −13/5 That gives us the point (3. We Eliminated. we must get the two original equations into SI form with t as the dependent variable. To begin. each followed by a step-by-step manipulation to get it into SI form with t as the dependent variable.”
Here’s a challenge!
Draw a graph of the above system with the variables transposed. If we plug in t = 3 to the SI version of the first equation.We Added. We don’t want any of the points to land off the scale on either axis. so we don’t have to drag ourselves through the justifications! 2t + 5u = −7 2t = −5u − 7 t = (−5/2)u − 7/2
. make u the independent variable and t the dependent variable. shown as a small open circle on the line for the second equation. We Can Graph
269
Let’s find another point on each line to make our line-drawing task easier. If we plug in t = 1 to the second equation. we get u = 4t − 3 =4×1−3 =4−3 =1 That gives us the point (1. shown as a small open circle on the line for the first equation. By now.
Solution
Let’s tackle this problem from scratch. Here are the original equations. a little common sense is a big help. Then we’ll make up a coordinate system with a horizontal u axis and a vertical t axis and plot the points. 1).
Are you confused?
When we want to find well-spaced points to draw lines in situations like this. but we have to get them far enough away from the other points so we can easily draw the lines. −13/5). That is. That will make it easy to find the t-intercepts.

t).3/4).3/4) u Solution = (–17/11. so we plot the intersection point as (−17/11. it must be of the form (u. The SI forms of the equations are shown below the originals.
. The solution is still t = 4/11 and u = −17/11. but when we write this as an ordered pair.270 Two-by-Two Linear Graphs
and u = 4t − 3 u + 3 = 4t 4t = u + 3 t = (1/4)u + 3/4 Now we know that the t-intercepts are −7/2 and 3/4. We’ve extended the negative (downward) t axis so we can plot the t-intercept for the first line and have a little extra room to extend the line past the point. On both axes. so we can plot the points corresponding to the ordered pairs (0. Figure 17-4 is a Cartesian graph of this situation.
t
u = 4t – 3 t = (1/4)u + 3/4
(0. each increment represents 1/2 unit.−7/2) and (0.4/11) Each axis increment is 1/2 unit (0. 4/11).–7/2)
2t + 5u = –7 t = (–5/2)u – 7/2
Figure 17-4 Graphs of 2t + 5u = −7 and u = 4t − 3
as a two-by-two linear system where the independent variable is u and the dependent variable is t.

Look again at the SI equation
w = 7v − 10 The slope of the line is 7.We Renamed. The other line passes through (0. so we can plot (2/3. −16/3). so we can plot (0. −5) on the coordinate plane. Let’s graph this system. If we go 2 units to the right. We Can Graph
271
We Renamed. −16/3).
Connect the points One line passes through (0. also called the substitution method: −7v + w + 10 = 0 and 4v + 8w = −40 We found that v = 2/3 and w = −16/3. calling v the independent variable and w the dependent variable. We Replaced. For the first equation:
−7v + w + 10 = 0 w + 10 = 7v w = 7v − 10 This tells us that the w-intercept for one of the lines is −10. 16. It ramps steeply upward as we move to the right. We Replaced. we’ll go 7 units upward. so we can plot the point (0. −10) on the Cartesian plane. These three known points are plotted in Fig.
Find two points for each line Our first step is to get both of the original equations into SI form with w as the dependent variable. We know that the lines intersect at the solution point where v = 2/3 and w = −16/3. The points here are badly “bunched up. For the second equation: 4v + 8w = −40 8w = −4v − 40 w = (−4/8)v − 40/8 w = (−1/2)v − 5 The w-intercept for the other line is −5. −5) and (2/3. we solved the following two-by-two linear system for v and w using the renameand-replace scheme.” Let’s find new points on both lines so we can obtain accurate graphs. for example. −10) and (2/3. we’ll go 14 units
. −16/3). We Can Graph
In Chap. If we go 1 unit to the right. 17-5.

In SI form. −5).–16/3) (0. The SI forms of the equations are shown below the originals. There’s no doubt that we need to find a more distant point on the line for the second equation. On both axes. How about 3 units to the right? Will the w value go off the scale? Let’s plug in v = 3 and see what we get: w = 7 × 3 − 10 = 21 − 10 = 11 It’s still in the field of view. that equation is w = (−1/2)v − 5 so the slope is −1/2.272 Two-by-Two Linear Graphs w (3. It passes through the point (0. 11) on the plane as a small open circle.–5)
Each axis increment is 2 units
(0.1) v Solution = (2/3. −10).–10) 4v + 8w = –40 w = (–1/2)v – 5
Figure 17-5 Graphs of −7v + w + 10 = 0 and
4v + 8w = −40 as a two-by-two linear system where the independent variable is v and the dependent variable is w. each increment represents 2 units. so we can plot (3. we can get a good idea of the position and orientation of the line.11) –7v + w + 10 = 0 w = 7v – 10 (–12. ramps gradually downward as
. and draw the line for the first equation through it and (0.
upward. Knowing this and the w-intercept.

at some point. this process takes only three steps. −5). Let’s plug in v = −12 and calculate: w = (−1/2) × (−12) − 5 =6−5 =1 This is in the field of view. so we can plot the point (−12. upward and to the right as well as downward and to the left.
Are you confused?
What happens when one of the lines in a graph has an undefined slope? How can you solve and graph a linear system such as this. and ramps gradually upward as we move to the left. Because this sloping line extends forever both ways. This is a perfectly decent two-by-two linear system. 17-5 by 90° (one-quarter of a turn) counterclockwise. we relabel the points by
. Now think: How can we morph this figure so the positive w axis goes toward the right and the positive v axis goes upward? If we can do that. you can solve it and graph it. we can do a trick with the coordinate system and the lines drawn on it. Then we mirror the drawing left-to-right. points. imagine Fig.” The positive v axis goes toward the right and the positive w axis goes upward. and draw the line for the second equation through it and (0. 17-5. We Replaced. 0).We Renamed. Finally. it must cross the vertical line x = 3. The original two equations are: −7v + w + 10 = 0 and 4v + 8w = −40 Suppose we want to draw a graph of this system with w as the independent variable and v as the dependent variable. Actually. 1) as a small open circle. 17-5 as an unbreakable unit with the axes. For extra credit. For a minute. We Can Graph
273
we move to the right. for example? y = 2x − 2 and x=3 You can’t reduce the second equation to SI form. and lines “all glued together. Its graph is a vertical line passing through (3. we’ll end up with a graph of the two-by-two linear system with w as the independent variable and v as the dependent variable. which extends forever straight upward and straight downward. We rotate the drawing in Fig. The first equation has a slope of 2. Instead of deriving the SI versions all over again with the variables reversed.
Here’s a trick!
Look again at the system we graphed in Fig.

) That transformation produces the graph of the inverse of a relation.
transposing the numbers in the ordered pairs. 17-5. as if we’ve done it before? We have! These maneuvers are the equivalent of mirroring the whole system along the axis corresponding to the line where the values of the independent variable and the values of the dependent variable are identical.0) (–5. and relabeling the points. Once we’ve done these three things. we have a graph of the two-by-two system with the variables transposed. 14. This graph was obtained by rotating Fig.3) w
(–10. mirroring it right-to-left. as you learned in Chap.0) Each axis increment is 2 units (1. Figure 17-6 shows the result.2/3)
–7v + w + 10 = 0 (11. For reference. here they are again: w = 7v − 10 and w = (−1/2)v − 5
. 17-5 by 90° counterclockwise. (In this case.–12)
Figure 17-6 Graphs of −7v + w + 10 = 0 and
4v + 8w = −40 as a two-by-two linear system where the independent variable is w and the dependent variable is v.274 Two-by-Two Linear Graphs v 4v + 8w = –40
Solution = (–16/3.
Are you astute?
Do you notice something familiar about the transformation we just performed. Look at the SI versions of the equations we graphed in Fig. that’s the line w = v.

We could not solve it.
. 6. 3) lies on the line for the first equation. Here’s the pair of equations: 2x + y = 3 and 6x + 3y = 12 Let’s graph these equations.
Find two points for each line If we consider x as the independent variable and y as the dependent variable.We Couldn’t Solve. then the SI version of the first equation is
y = −2x + 3 and the SI version of the second equation is y = −2x + 4 The y-intercepts are 3 and 4. and 7 at the end of Chap. Here are the manipulations. This will help us see why no ordered pair (x. starting with the original equations: −7v + w + 10 = 0 −7v + 10 = −w −7v = −w − 10 7v = w + 10 v = (1/7)w + 10/7 and 4v + 8w = −40 4v = −8w − 40 v = (−8/4)w − 40/4 v = −2w − 10 These inverse relations are both functions of w. If we put these equations into SI form with v rather than w as the dependent variable.y) satisfies them as a system. but We Can Graph
In exercises 5. That means the point (0. respectively. and the point (0. but We Can Graph
275
These are both functions of v. we scrutinized a two-by-two linear system. 4) lies on the line for the second equation. we obtain the inverse relations. Can you see why?
We Couldn’t Solve. 16.

The lines are parallel and distinct. we’ll end up at (4. each increment represents 1 unit. That means we must go downward by 8 units. taken together. These two lines. −5). If we move n units to the right along either line (where n is any number). On both axes.
.–5)
Figure 17-7 Graphs of 2x + y = 3 and 6x + 3y = 12
as a two-by-two linear system where the independent variable is x and the dependent variable is y. Starting at (0. 3). so we can’t use their point of intersection as a graph-plotting aid.–4) (4.3) 6x + 3y = 12 y = –2x + 4 x
Each axis increment is 1 unit
(4. Note that the lines are parallel. we must move 2n units downward to stay on the line.276 Two-by-Two Linear Graphs
When taken together as a two-by-two linear system. The SI forms of the equations are shown below the originals. 4).4) (0. Both lines have slopes of −2. we’ll end up at (4. form the graph of the two-by-two linear system represented by the inconsistent equations
2x + y = 3 and 6x + 3y = 12
y 2x + y = 3 y = –2x + 3 (0. so they have no intersection point. Let’s move to the right by 4 units along each line.
Connect the points Figure 17-7 shows the four points we’ve found. But there’s another way. That means their graphs do not intersect. and the two lines connecting them. −4). these equations have no common solution. Starting at (0.

then they both have slopes of 0. and they both represent functions. horizontal lines in the same coordinate system. although they are both relations between x and y. Label a few of the infinitely many ordered pairs that satisfy the system.We Couldn’t Solve. then neither of the equations is a function of x. so if we start at any point on the line and move to the right by 1 unit (add 1 to r). State and graph that function. and a ≠ b.
Here’s a challenge!
In exercise 10 at the end of Chap. 16. The slope is 2. Additional points can be found by moving to the right or left from (0. It tells us that the s-intercept is −3. but We Can Graph
277
Are you confused?
Suppose the graphs of a two-by-two linear system show up as parallel.
Solution
The function can be stated as the first equation above. vertical lines. we tried to solve the following pair of equations as a linear system by substitution. and c ≠ d. so we can plot (0. Such a system can always be reduced to the form y=c and y=d where c and d are constants. If the graphs of a two-by-two linear system show up as parallel. we must move upward by 2 units (add 2 to s) to stay on the line. If you call x the independent variable and y the dependent variable. −3) and moving upward or downward by twice that distance. but failed when we got the meaningless result 0 = 0: s = 2r − 3 and −10r + 5s + 15 = 0 We showed that these equations are equivalent to a single function of r. both with undefined slope. −3) on the Cartesian plane. we must move downward by 2 units
. This sort of system can always be reduced to the form x=a and x=b where a and b are constants. letting r be the independent variable and s be the dependent variable. If we move to the left by 1 unit (subtract 1 from r).

3) (2.5) (3. You may (and should) refer to the text as you solve these problems. If you think you can solve a particular problem in a quicker or better way than you see there. by all means try it!
.–1) Solutions = all points on line (–1.278 Two-by-Two Linear Graphs s s = 2r – 3 (4.
(subtract 2 from s). but nondenumerable!
Practice Exercises
This is an open-book quiz. along with the line connecting them. each increment represents 1 unit. B. Don’t hurry! You’ll find worked-out answers in App. The line is the graph of the function s = 2r − 3 Any point on the line (not only the ones plotted in the figure) can be called a solution to this redundant two-by-two linear system. the solution set (the set containing all the ordered pairs that satisfy the system) is not only infinite.1) r (1. The solutions in the appendix may not represent the only way a problem can be figured out. On both axes. The lines coincide because the two equations are equivalent. 17-8.–3)
Each axis increment is 1 unit
Figure 17-8 Graphs of s = 2r − 3 and −10r + 5s + 15 = 0
as a two-by-two linear system where the independent variable is r and the dependent variable is s.–5) –10r + 5s + 15 = 0 (0. Some of the resulting points are plotted in Fig. Because the domain and the range are both the entire set of real numbers.

show that the solution to the equations derived in Probs. Using the graph derived in Prob. f (x) is just another name for y. In this context. 8.
1. Refer to Fig. Call the transposed line L by the new name L*. 4.Practice Exercises y 6 Solution = (–3. Refer to Fig.4)
279
L
M
Figure 17-9 Illustration for Practice Exercises 1 and 2. 17-9. which shows up as the point (−3. what is the equation of line L* in SI form? Remember that x is now the dependent variable. What is the equation of line L in SI form? 2. obtaining a new graph that shows the system with y as the independent variable and x as the dependent variable. show that the solution to the equations derived in Probs. 6. and call the transposed line M by the new name M*. 17-9. Using the graph derived in Prob. 0). 4. what is the equation of line M* in SI form? Remember that x is now the dependent variable. Suppose we see a linear function where x is the independent variable and y is the dependent variable. Now imagine that we derive the inverse of this function so y
. Rotate and mirror Fig. Let’s call the function f and state it like this: f (x) = mx + b where m is the slope and b is the y-intercept of the graph. 7. and x = −3. Using the morph-and-mix method. 4. −3).0) 4 2 x –6 (0. 5 and 6 is y = 0. which shows up as the point (0. 1 and 2 is x = −3 and y = 0. Using the morph-and-mix method. What is the equation of line M in SI form? 3. 5.–2) –4 –6 2 4 6 (0. 17-9.

(Some texts will say.) Under what conditions is the inverse relation of a linear function not another function? Here’s a hint: A straight line in Cartesian coordinates represents a function if and only if its slope is defined. such as this book.280 Two-by-Two Linear Graphs
becomes the independent variable and x becomes the dependent variable. but that contains the constants m and b instead of the constants n and c. in place of the isolated x. then f always has an inverse relation. under such conditions. Here’s a hint: state f as y = mx + b and morph this into SI form with x alone on the left side of the equals sign. or that f −1 does not exist. 10. Others. Derive a version of f −1 as we defined it in Prob. but not functions.
. Then. and state it as follows: f
−1
( y) = ny + c
where n is the slope and c is the x-intercept. 8. write f −1( y). will say that if we consider f as a relation. but that relation might not be a function. Draw graphs of three linear functions in Cartesian coordinates whose inverses are relations. f −1 is almost always a function. Now if f is a function. 9. We call this inverse function f −1. But sometimes it isn’t. that f has no inverse.

and d are constants. for reference:
−4x + 2y − 3z = 5 and 2x − 5y − z = −1
. c. The first equation is already in this form. b. we must get all three equations into the same form.282 Larger Linear Systems
Get the equations into form Now that we’ve chosen the variable to eliminate. The second equation is 2x − 5y = z − 1 Subtracting z from each side will put it into the form we want: 2x − 5y − z = −1 The third equation is 3x = −6y + 7z Adding 6y to each side. we get 3x + 6y = 7z We can subtract 7z from each side and it comes into the sought-after form: 3x + 6y − 7z = 0 We now have the three-by-three system in this uniform condition: −4x + 2y − 3z = 5 2x − 5y − z = −1 3x + 6y − 7z = 0
Make z vanish once Here are the first two revised equations again. The following form is as good as any:
ax + by + cz = d where a.

we decided to work on the first two equations in the revised three-by-three system. we get −14x + 35y + 7z = 7 3x + 6y − 7z = 0 −11x + 41y = 7
State the two-by-two We have now derived two different equations in the variables x and y. and then between the first equation and the third
. This is a two-by-two linear system. and then work on the second two equations. getting −6x + 15y + 3z = 3 If we add this to the first equation.Eliminate One Variable
283
Let’s multiply the second equation through by −3. “Why can’t we eliminate z between the first equation and the second one. Here they are again:
2x − 5y − z = −1 and 3x + 6y − 7z = 0 We can multiply the first equation through by −7. getting −14x + 35y + 7z = 7 When we add the second equation to this. You might ask. we obtain the sum −4x + 2y − 3z = 5 −6x + 15y + 3z = 3 −10x + 17y = 8
Make z vanish again Now let’s scrutinize the second two revised equations. which we know how to solve:
−10x + 17y = 8 and −11x + 41y = 7
Are you confused?
When we set out to get rid of the variable z.

As you continue to study algebra and take more advanced courses.
Here’s a challenge!
Put the original equations in our three-by-three linear system into a form where z appears all by itself on the left sides of the equals signs. and y appear on the right sides. This defines two independent variables (in this case x and y) and a single dependent variable (z). y) = 2x − 5y + 1 h (x. You have learned enough algebra so you can follow along without detailed explanations. x. and expressions containing only constants. You might also learn techniques to solve four-by-four or larger systems. If we call the relations f. In any case. y) = (−4/3)x + (2/3)y − 5/3 g (x.284 Larger Linear Systems
one? Or between the first and the third. −4x + 2y − 3z = 5 2y − 3z = 4x + 5 −3z = 4x − 2y + 5 −z = (4/3)x − (2/3)y + 5/3 z = (−4/3)x + (2/3)y − 5/3 2x − 5y = z − 1 2x − 5y + 1 = z z = 2x − 5y + 1 3x = −6y + 7z 3x + 6y = 7z (3/7)x + (6/7)y = z z = (3/7)x + (6/7)y These three equations represent relations that map pairs of variables (in this case x and y) into a single variable (in this case z). g.” Whenever you want to solve a three-by-three system of equations. “We can! Either of those alternative schemes will work just as well as the one we chose. for reference: −4x + 2y − 3z = 5 2x − 5y = z − 1 3x = −6y + 7z
Solution
To morph these equations. but the ultimate solution to the three-by-three system will turn out the same. step-by-step. then we can write: f (x. you’ll learn other ways to solve three-by-three systems than the methods presented in this book. Here are the processes. and h. you must give every one of those equations some “say” in the outcome. you must somehow involve all three of the equations in the solution process. Here are the original equations. we can use the same sort of algebra that gets two-variable linear equations into SI form. y) = (3/7)x + (6/7)y
. The intermediate equations will differ. if you want to find the solution to a system of equations. and then between the second and the third?” The answer is.

Here it is again: −10x + 17y = 8 and −11x + 41y = 7 We can use any of the methods described in Chap.Solve the Two-by-Two
285
Solve the Two-by-Two
Let’s continue our quest to solve the three-by-three linear system. we get
110x − 187y = −88 and −110x + 410y = 70 Now let’s add these two equations in their entirety: 110x − 187y = −88 −110x + 410y = 70 223y = −18 Dividing through by 223 tells us that y = −18/223. Let’s use double elimination. We have obtained a two-bytwo system in x and y. solve for y Let’s multiply the first equation through by −11 and the second equation through by 10.
Eliminate x .
Eliminate y . 16 to tackle this. That gives us
410x − 697y = −328 and −187x + 697y = 119
. This fraction happens to be in lowest terms (a fact that you can verify if you like). solve for x Let’s multiply the first equation in our two-by-two system through by −41 and the second equation through by 17. When we do that.

This fraction. like the previous one with the same denominator.”
Here’s a challenge!
Solve the preceding two-by-two system using the morph-and-mix method.
Two down. We have the values for x and y :
x = −209/223 and y = −18/223 In the next section.” you might ask. is in lowest terms. of course. Consider x the dependent variable. here’s our pair of equations: −10x + 17y = 8 and −11x + 41y = 7 We must get both of these into SI form. the first equation morphs like this: −10x + 17y = 8 −10x = −17y + 8
. Then we’ll check our work. Something tells me that z is going to be a fraction with a denominator of 223. “Why. we get 410x − 697y = −328 −187x + 697y = 119 223x = −209 Dividing through by 223 tells us that x = −209/223. “We can use either of those methods. What do you think?
Are you confused?
You might again question the choice of solution processes.286 Larger Linear Systems
Adding these equations.
Solution
Once again. with x all by itself on the left sides of the equals signs. we’ll substitute these values back into one of the original three equations and solve for z. “do we use the doubleelimination method to solve the two-by-two system here? Why not use morph-and-mix or rename-andreplace?” The answer is. Step-by-step. one to go! We’ve now solved for two of the three unknowns in our three-by-three system.

784)/2. we have −223y = 18 We can divide through by −223 to get y = 18/(−223) = −18/223. The first equation will do. and multiply the numerators and denominators on the right side of the equals sign by 10. This agrees with the result we obtained by double elimination.230 = (−306 − 1. solving for x.230 − 8/10 = −306/2.784/2. We can multiply the numerators and denominators on the left side of the equals sign by 11.230 = −209/223
. We can plug this into either of the SI equations we derived above. we obtain 187y − 88 = 410y − 70 Adding 88 to each side produces 187y = 410y + 18 Subtracting 410y from each side. Step-by-step: x = (17/10)y − 8/10 = (17/10)(−18/223) − 8/10 = −306/2. That gives us (187/110)y − 88/110 = (410/110)y − 70/110 Multiplying the whole equation through by 110 to get rid of the fractions.230 = −2.090/2.230 − 1. getting a first-degree equation in y : (17/10)y − 8/10 = (41/11)y − 7/11
287
Let’s get a common denominator here.Solve the Two-by-Two
10x = 17y − 8 x = (17/10)y − 8/10 The second equation morphs as follows: −11x + 41y = 7 −11x = −41y + 7 11x = 41y − 7 x = (41/11)y − 7/11 Now we mix the right-hand sides.

If we made an error somewhere. Let’s use the third original equation:
3x = − 6y + 7z Plugging in the numbers for x and y. we get 3 × (−209/223) = −6 × (−18/223) + 7z −627/223 = 108/223 + 7z −735/223 = 7z 7z = −735/223 z = −105/223
Are you confused?
If the last step in the above calculation confuses you. at least tentatively: x = −209/223 y = −18/223 z = −105/223
. and they’ve come out the same both times. note that −735/7 = −105. By now. and we’re confident that they’re correct because we’ve arrived at them from two different directions. because I acted on my hunch that z would be a fraction with a denominator of 223. That’s the numerator in the fraction. and proceeding step-by-step. but it would be a distraction to delve into the “why” of it right now. you have probably noticed that in linear systems. Here they are again: x = −209/223 and y = −18/223
Plug them in We can use any of the original equations or their revisions to solve for z. I had a feeling that −735 would cleanly divide by 7. We’ve found x and y by two different routes. the answers would almost certainly disagree. fractional solutions have a way of coming out with identical denominators. The important thing is that we have all three solutions to our original three-by-three linear system.288 Larger Linear Systems
It agrees again! This practically guarantees that our answers are correct. There’s a good reason for that.
Substitute Back
Now we have the values for x and y.

3). It makes use of three coordinate axes. z). plotted in Cartesian three-space. as you saw in Chap. and the coordinates of point Q are (3. their graphs can show up in any of the following ways:
• Three different planes that all intersect at a single. the second number represents the value on the y axis. P and Q. y. In the second case. as in Fig. 17. “What does a three-by-three graph look like?” or “What happens when there are many linear equations in many variables?” or “What happens when the number of equations is not the same as the number of variables?”
Two-by-two geometry The graph of a linear equation in two variables shows up as a straight line in the Cartesian plane. When you have two such equations. corresponding to the point where the lines intersect. In the third situation. 18-1. unique point • Two different lines that are parallel • Two lines that precisely coincide In the first case. and the third of which intersects the other two in a pair of parallel lines. two of which are parallel. If the drawings were literal.
. −2).290 Larger Linear Systems
General Linear Systems
Now that you’ve seen two-by-two and three-by-three linear systems. the x axis would appear horizontal on the page. Three-by-three geometry The graph of a linear equation in three variables appears as a flat plane (not a line!) in Cartesian three-space. there are infinitely many solutions. the y axis would appear vertical on the page. 5. Note that the positive x axis goes to the right. the positive y axis goes upward. In this example. and the third number represents the value on the z axis. where the first number represents the value on the x axis. The most common such system is called Cartesian three-space. and in such a way that each axis is perpendicular to the other two. the variables are x. The coordinates of point P are (−5. Cartesian three-space is sometimes drawn in perspective. all of which intersect at their zero points. −4. we need to use a system that can portray all of space. • Three different planes. and the z axis would be perpendicular to the page. and z. Instead. the system has a single solution. unique point. you might wonder. their graphs always appear in one of three ways:
• Two different lines that intersect at a single. Points are denoted as ordered triples in the form (x. y. Figure 18-2 shows two specific points. When we have three linear equations in three variables. there is no solution. and the positive z axis comes toward us.
Cartesian three-space The graph of a linear equation in three variables can’t be drawn on a Cartesian plane.

The x axis increases positively
from left to right.5. All three axes intersect at the origin and are mutually perpendicular.+y
–z –x +x
+z –y
Figure 18-1 Cartesian three-space. each increment represents 1 unit. On all three axes.–4.y.3) –y
Figure 18-2 Two points in Cartesian three-space. 291
.z).–2)
+y
Each axis increment is 1 unit –z –x +x
+z P (–5.
Q (3. the y axis increases positively from the bottom up. and the z axis increases positively from far to near. along with the
corresponding ordered triples of the form (x.

When n is large. if we write up a system of n linear equations in n variables “at random. but not always. A true view would require a three-dimensional hologram that we could walk around in! When it comes to n-by-n linear systems where n is a natural number larger than 3. There are infinitely many such points. 16! As n increases. give only a limited perspective. and such programs can be a big help. a unique solution exists in this type of system. • Three different planes that are mutually parallel. and a third plane parallel to them both. • Two planes that precisely coincide. Even so. then it has infinitely many solutions. and z).y. Computer programs have been developed to portray systems like these. there is no single solution to the whole system. The process of solving an n-by-n linear system is ideally suited to computer applications. In the Cartesian plane. Try to envision all these situations. corresponding to the point where all three planes intersect. Often. • Three planes that all precisely coincide. In each of the second through fifth cases.292 Larger Linear Systems
• Three different planes. In the first case. there are many ways that an n-by-n linear system can fall short of a unique solution. so does the time it will take us to solve the system. Think of a linear system with two variables but only one equation. two variables).z) that represent points on the plane. Note that two planes are parallel in space if and only if they do not intersect. unless we have access to a computer.
. As you can imagine. there are infinitely many solutions. This type of system can never have a unique solution. the process of solving an n-by-n system of linear equations where n > 3 is bound to be time-consuming and tedious. the system has a unique solution. and all three of those lines are mutually parallel. and a third plane that intersects them both in a single line. In each of the last three cases. too. A unique solution always comes down to a single point in Cartesian n-space. See how much longer it took us to solve the three-by-three system in this chapter than it took us to solve the two-by-two systems in Chap. Now imagine a one-by-three linear system (a single equation in x. It’s not easy to verbally describe what happens in n-by-n linear systems when n is large. a linear system has fewer equations than variables. even a walk-through hologram can’t give us a complete picture. Whenever that happens. it shows up as a single straight line. But they. there is no solution. which grind out solutions by brute force. • Three different planes that all intersect in a single line.
Fewer equations than variables All of the linear systems we’ve examined so far have the same number of equations as variables. Its graph in Cartesian three-space is a single flat plane. Occasionally. If we consider this as a one-by-two linear system (one equation. such that each pair of planes intersects in a different line. y. so this type of system has infinitely many solutions.
Are you confused?
Most people have trouble envisioning the graphs of three-by-three linear systems. • Two planes that precisely coincide. The solutions are all the ordered triples (x. It is the same thing as a redundant two-by-two system.” the chance is good that it will have a single solution.

the point can be represented by a unique ordered triple. and −3. then adding one or more extra equations cannot make it consistent. but there is no single point common to all three lines. Then they don’t intersect anywhere. If there are n variables.General Linear Systems
293
Things get more interesting with two-by-three linear systems (two equations and three variables x. These statements can be generalized to any number of linear equations in any number of variables.” We can imagine such a system as a two-by-two or three-by-three linear system with one or more extra equations thrown in. no matter how many variables there are.
Here’s a challenge!
Draw a graph of the following three-by-two linear system to illustrate why it does not have a unique solution: y=x+1 y = 3x − 1 y = −2x − 3
Solution
These three equations are in SI form.
More equations than variables When a linear system in two variables has more than two equations. The y-intercepts are 1. If a two-by-two or three-by-three linear system is consistent. and z). If there are two variables. That means a two-by-three linear system can’t have a unique solution. Then. but not necessarily. Two flat planes in space can never intersect in a single point.z) that correspond to points on that line.y. The slopes are 1. we can be certain that the system has no unique solution. that point can be represented by a unique ordered pair. which portrays the graphs of the equations as a single system. If a two-by-two or three-by-three linear system is inconsistent. 3. and there are no solutions to the system. the algebra will lead us into absurd or useless statements. y. we have “extra data. The graph appears as two planes in Cartesian three-space. respectively. again.y. or a linear system in three variables has more than three equations.z) that correspond to points on that plane. preferably with the help of a computer. If we encounter 27 equations in 28 variables. we can’t know if the system has a unique solution until we try to solve it.
. the graphs of all the equations must have a single point in common. If the system has no unique solution. For any linear system to have a solution. 18-3. they can be used to find second points as shown in Fig. A second possibility is that the two planes are parallel. and −2 respectively. Each line intersects both of the others. the point can be represented by a unique ordered n-tuple. The planes might intersect in a straight line. But if we see 28 or more equations in 28 variables. then adding one or more extra equations might make it inconsistent or redundant. A third possibility is that the planes coincide. in which case there are infinitely many solutions: all the ordered triples (x. −1. If there are three variables. there are infinitely many solutions: all the ordered triples (x.

The solutions in the appendix may not represent the only way a problem can be figured out.5)
(–3. There is no solution. You may (and should) refer to the text as you solve these problems.–1) (0.5) (4.” we got rid of z between the first two of these equations. Now eliminate z between the first and third equations. Here are the three revised original equations for the three-by-three system we tackled in this chapter: −4x + 2y − 3z = 5 2x − 5y − z = −1 3x + 6y − 7z = 0 In the section “Eliminate One Variable.1) x (0. because there is no single point common to all three lines.
Practice Exercises
This is an open-book quiz. each increment represents 1 unit. Don’t hurry! You’ll find worked-out answers in App. If you think you can solve a particular problem in a quicker or better way than you see there.
. by all means try it! 1.–3) Each axis increment is 1 unit
y=x+1 y = 3x – 1
Figure 18-3 Graphs of three equations in two variables.
considered as a linear system. On both axes.294 Larger Linear Systems y y = –2x – 3
(2. B. and then between the second two.3) (0.

Solve for z by substituting the values for x and y (solution 6) into the second equation stated in Prob. Solve for z by substituting the values for x and y (solution 3) into the first equation stated in Prob. Graph all four of the lines presented in Prob. 7. explain why any pair or triplet of these equations. along with the equation in x and y that we derived from the first two three-variable equations in the section “Eliminate One Variable. 5.
. Solve the two-by-two linear system obtained in the solution to Prob. has the same unique solution as any other pair or triplet of the equations. On the basis of this graph. The following four-by-two linear system has a unique solution. 5. Use the morphand-mix method. 10. 1. taken as a two-by-two or three-by-two system. 8.Practice Exercises
295
2. 1. along with the equation in x and y that we derived from the second and third three-variable equations in the section “Eliminate One Variable. Plug in the values for x and y that appear to solve the set of equations in Prob.” 3. even though there are more equations than variables. 8. Derive a two-by-two linear system in x and y from the solution to Prob.” 6. based on the reasoning in the solution to Prob. Verify that these values satisfy all four equations. 4. How can we know this without doing any algebra or graphing the equations? What is that solution? y = −x + 1 y = −2x + 1 y = 3x + 1 y = 4x + 1 9. 2. treating y as the independent variable and x as the dependent variable. 1. 1. Use the doubleelimination method. 8. Derive a two-by-two linear system in x and y from the solution to Prob. Solve the two-by-two linear system obtained in the solution to Prob. 8.

“Can matrices represent two-by-two systems? Can they represent systems larger than threeby-three? Can they represent asymmetrical systems such as five-by-four?” The answers are “Yes. we’ll look at three-by-three systems only. For the first equation: 7y = 3z + 3 0x + 7y = 3z + 3 0x + 7y − 3z = 3 For the second equation: 8z = −2x − 7 2x + 8z = −7 2x + 0y + 8z = −7 For the third equation: 12x = 7y 12x − 7y = 0 12x − 7y + 0z = 0
.” “Yes. we can remove the variables and equals signs. But they aren’t really necessary.
Are you confused?
You might ask.
Here’s a challenge!
Put the following three-by-three linear system into matrix form: 7y = 3z + 3 8z = −2x − 7 12x = 7y
Solution
None of these three equations is in the proper form for conversion to matrix notation. each with four numbers:
a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d3
Some texts enclose matrices in huge parentheses or brackets. but matrix techniques can be applied to any linear system. Let’s not use them.” and “Yes.How to Build a Matrix
297
The matrix form Once we have morphed all the equations into the proper form. We’ll have to manipulate them. step-by-step. Here are the processes. Then we can put the numbers into an array having three rows. and they can clutter things up.” In this chapter.

if we start with
a1 a2 a3 we can change it to a3 a2 a1 b3 b2 b1 c3 c2 c1 d3 d2 d1 b1 b2 b3 c1 c2 c3 d1 d2 d3
In this case. the first and third rows have been swapped. multiply. We can make as many of these moves as we want. arranged in three horizontal rows and four vertical columns. Note that we cannot swap individual elements or vertical columns! The swap maneuver is only allowed between entire rows. Let’s invent a matrix morphing game.
Swap We may interchange all the elements between two rows in a matrix.298 The Matrix Morphing Game
Now we have these three equations that make up the linear system: 0x + 7y − 3z = 3 2x + 0y + 8z = −7 12x − 7y + 0z = 0 We may want to write the above equations like this. and add. so we are sure to get the signs of the coefficients right: 0x + 7y + (−3z) = 3 2x + 0y + 8z = −7 12x + (−7y) + 0z = 0 We can write this system in matrix form by removing the variables and equals signs.
. while keeping the elements of both rows in the same order from left to right. There are three types of moves in this game: swap. For example. and then aligning the coefficients into neat rows and columns: 0 2 12 7 0 −7 −3 8 0 3 −7 0
Matrix Operations
Imagine the matrix for a three-by-three linear system as a game board with 12 positions.

all the elements in the second row have been multiplied by k. As with the swap move. we can operate only on entire rows. Because the absolute value of k can be smaller than 1. For example. if we have
a1 a2 a3 we can change this to a1 ka2 a3 b1 kb2 b3 c1 kc2 c3 d1 kd2 d3 b1 b2 b3 c1 c2 c3 d1 d2 d3
In this case. keeping the elements in the same order from left to right. taking care to keep the elements of both rows in the same order from left to right. and then replace the elements in either row by the sum. we aren’t allowed to replace the third row with the sum of the first and second rows.Matrix Operations
299
Multiply We may multiply all the elements in any row by a nonzero constant. We can’t do this maneuver with individual elements or with columns. In this example.
Add We may add all the elements in any row to all the elements in another row. we can extrapolate this rule to let us multiply or divide all the elements in any row by a nonzero constant.
. Suppose we start with this matrix:
a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d3
We can change it to either of the following:
a1 a1 + a2 a3 or a1 + a2 a2 a3 b1 + b2 b2 b3 c1 + c2 c2 c3 d1 + d2 d2 d3 b1 b1 + b2 b3 c1 c1 + c2 c3 d1 d1 + d2 d3
Note that the replaced row must be one of the two involved in the sum.

you are right.
A Sample Problem
In this section. no matter what road we take!
The game plan Anyone who plays a game needs a plan. has an ultimate objective. provided the system is consistent (has a unique solution). First. But such a move would delete the information in the equation represented by the replaced row. We would be left with the equivalent of a two-equation system in three variables. we’ll solve a three-by-three linear system using matrix operations. make it into a matrix. Such a system doesn’t contain enough information to define a unique solution. leaving us with a two-equation system having three variables. and then play the matrix morphing game until we get the unit diagonal form. and z. and z are real numbers. will represent the solution to the linear system? If so. Do you suspect that the values x. This is called the unit diagonal form. we could try it.
Are you still confused?
You might also ask. The process in this section doesn’t represent the only avenue by which the final result can be reached. turning it into a mere hybrid of the other two rows. y. when in fact it was consistent. we get equations into form:
a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3
. Our goal is to get a matrix representing a three-by-three linear system into this form:
1 0 0 0 1 0 0 0 1 x y z
where x. and for awhile it might seem to work. Here’s a strategy for the matrix morphing game that works well for me. The game can be played in many ways. Doing that would wipe out one of the equations in the system. Now imagine that we start with a three-by-three linear system. like any sensible game. We might end up thinking that the original system was redundant. which appear in the far right column. we’ll always get to the same destination. “Why can’t we add two rows in a matrix and then replace the remaining row (the one not involved in the sum) with that sum?” Well. But if we manage to avoid making mistakes.
Are you confused?
Do you wonder why we can’t multiply all the elements in a row by 0? Think about this for a minute. y.300 The Matrix Morphing Game
The final goal The matrix morphing game.

and an at sign (@) can represent any nonzero real number. we may find that one or more of the solutions x. our failure indicates one of four things:
• • • • We didn’t try hard enough We made a mistake somewhere The original system is inconsistent The original system is redundant
The system to be solved Now let’s methodically tackle a three-by-three linear system and solve it using the matrix morphing game.
What if we can’t “win”? If we find it impossible to get a matrix into the unit diagonal form. we get the matrix into echelon form.A Sample Problem
301
Then we make the matrix:
a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d3
Next. which looks like this:
# 0 0 # # 0 # # # # # #
where a pound sign (#) can represent any real number. or z is a fraction that can be reduced. We ought to reduce all solutions to their lowest forms in the interest of elegance. Our final goal is the unit diagonal form:
1 0 0 0 1 0 0 0 1 x y z
Once we’ve put a matrix into the unit diagonal form. Here are the equations:
3x + z = 2y + 11 4y + 2z = x −5x + y = 3z − 20
. y. Then we go for the diagonal form:
@ 0 0 0 @ 0 0 0 @ # # #
where a pound sign can represent any real number.

302 The Matrix Morphing Game
Formatting the equations None of these equations is in the proper form for assembling a matrix. we can subtract 2y from each side to get
3x − 2y + z = 11 With the second equation. Let’s get them that way! Fortunately. and can state the whole system like this: 3x − 2y + z = 11 −x + 4y + 2z = 0 −5x + y − 3z = −20
Building the matrix To construct the matrix. We can multiply the second row by −5 to obtain
3 5 −5 −2 −20 1 1 −10 −3 11 0 −20
We can add the second and third rows and then replace the third row with the sum to get
3 5 0 −2 −20 −19 1 −10 −13 11 0 −20
. With the first equation. paying close attention to the signs:
3 −1 −5 −2 4 1 1 2 −3 11 0 −20
Deriving the echelon form There are many different routes by which we can arrive at an echelon form of this matrix. getting −5x + y − 3z = −20 We now have the equations in form. the maneuvers are simple. Let’s start by getting 0 at the extreme left in the bottom row. we remove all the variables and arrange the remaining numbers into an orderly array. we can subtract 3z from each side. we can subtract x from each side to obtain −x + 4y + 2z = 0 With the third equation.

we get
3 0 0 −2 190 −190 1 133 −130 11 209 −200
Adding the second and third rows and then replacing the third row with the sum. we have
3 0 0 −2 10 −19 1 7 −13 11 11 −20
Now the number −19 in the third row must somehow be made to vanish. we get
3 0 0 −2 190 0 0 133 −1 8 209 −3
. which means we must turn it into 0. In this case.A Sample Problem
303
Now let’s get 0 at the extreme left in the second row. 1. Let’s use the second row to “attack” it. If we multiply the second row by −3/5. we want to get 0s in the places that now contain −2. Let’s divide the third row by −3 to get
3 0 0 −2 190 0 1 133 −1 11 209 −3
Adding the first and third rows and then replacing the first row with the sum. we obtain a matrix in echelon form:
3 0 0 −2 190 0 1 133 3 11 209 9
Deriving the diagonal form To morph the echelon matrix into diagonal form. we come up with
3 −3 0 −2 12 −19 1 6 −13 11 0 −20
Adding the first and second rows and then replacing the second row with the sum. We can use the third row to “attack” it. currently equal to 1. and 133. we simply keep playing the game. If we multiply the second row through by 19 and the third row through by 10 (combining two moves). Let’s start with the third element in the first row.

We can divide the top row by 3 and the middle row by 2. obtaining the unit diagonal matrix
1 0 0 0 1 0 0 0 1 2 −1 3
. we get
3 0 0 −2 190 0 0 133 −133 8 209 −399
Adding the second two rows and then replacing the second row with the sum. and divide the third row through by −133.304 The Matrix Morphing Game
Let’s make the number 133 vanish. we obtain
3 0 0 −2 190 0 0 0 −133 8 −190 −399
We now have an opportunity to reduce the sizes of the numbers in the second and third rows. Let’s divide the second row through by 190. we get
3 0 0 0 2 0 0 0 1 6 −2 3
Deriving the unit diagonal form Our remaining task is simple and clean. We can multiply the second row by 2 to obtain
3 0 0 −2 2 0 0 0 1 8 −2 3
Adding the first and second rows and then replacing the first row with the sum. That gives us
3 0 0 −2 1 0 0 0 1 8 −1 3
We have only to make the element −2 in the first row vanish. We can use the third row as the “weapon” here. and we’ll have the matrix in diagonal form. If we multiply the third row by 133.

because it’s easy to make mistakes in the matrix morphing game. Finally. Mission accomplished. We must check our work! Only then can we be totally confident that our solution is correct. (Otherwise we’d have fractions with large numerators and denominators.
x=2 y = −1 z=3
Are you confused?
If you think we’ve finished solving this problem. you’re indeed confused. but take a different route this time. We have no fractions to reduce. Don’t reduce the sizes of any of the integers.A Sample Problem
305
Stating and reducing the solution The solution to the system is now clear. When we plug the values for x. Let the absolute values grow as large as they “want”!
. the third equation: −5x + y = 3z − 20 −5 × 2 + (−1) = 3 × 3 − 20 −10 − 1 = 9 − 20 −11 = −11 Check three. and z into the first original equation. although they would divide out cleanly and leave us with integers. y.) Evidently.
Here’s an extra-credit challenge!
Play the matrix morphing game with the linear system we just got done solving. because we kept the sizes of the numbers down as we went along. here’s what happens: 3x + z = 2y + 11 3 × 2 + 3 = 2 × (−1) + 11 6 + 3 = −2 + 11 9=9 Check one. Now for the second equation: 4y + 2z = x 4 × (−1) + 2 × 3 = 2 −4 + 6 = 2 2=2 Check two.

you should end up with sloppy fractions in the far-right column. 4 into diagonal form. 5 to a form with the smallest possible absolute values in each row. When you reduce these fractions. you should get 1 0 0 0 1 0 0 0 1 2 −1 3
Solution
You’re on your own. Reduce the matrix derived in the solution to Prob.
. Put the matrix derived in the solution to Prob. 8. 6. such that all the numbers in the matrix are integers. plug the numbers into the equations stated in the solution to Prob. Put the matrix of Prob. Write the set of equations from the solution to Prob. 3 into echelon form. 7 to be sure they’re correct. Put the following three-by-three linear system into the proper form for conversion to matrix notation: x=y−z−7 y = 2x + 2z + 2 z = 3x − 5y + 4 2. To do this. 1 in the form of a matrix. 3 and stated in solution 3. 7. y. Write the set of equations represented by the following matrix:
0 5 1 4 −3/2 1 −1 8 1 −2 1 1
4. by all means try it! 1. Then state the tentative solution to the three-by-three linear system we derived from the matrix in Prob. Check the values for x. and z derived in the solution to Prob. 5.306 The Matrix Morphing Game
When you get the unit diagonal form. The solutions in the appendix may not represent the only way a problem can be figured out. Don’t hurry! You’ll find worked-out answers in App. 3. 3. You may (and should) refer to the text as you solve these problems. Reduce the matrix derived in the solution to Prob. Have fun!
Practice Exercises
This is an open-book quiz. If you think you can solve a particular problem in a quicker or better way than you see there. B. 6 to unit diagonal form.

Put the following three-by-three linear system into matrix form. trying to get the unit diagonal form. It can’t be done! Why? x+y+z=1 2x + 2y + 2z = 2 3x + 3y + 3z = 3
. trying to get the unit diagonal form.Practice Exercises
307
9. Then play the matrix morphing game for awhile. Put the following three-by-three linear system into matrix form. It can’t be done! Why? x+y+z=1 x+y+z=2 x+y+z=3 10. Then play the matrix morphing game for awhile.

we write p ≠ q. we write p ≥ q.
Question 11-4
Suppose we divide an inequality through by a nonzero real number. then we can also say x is smaller than or equal to y. There are infinitely many other examples. Under what circumstances will the sense of the inequality be reversed? Under what circumstances will the sense of the inequality stay the same?
Answer 11-4
The sense of the inequality is reversed if we divide through by a negative real number. If p is not equal to q but we don’t know which is smaller. If p is smaller than or equal to q.” The above statement can be written symbolically as (x < y) ⇒ (x ≤ y)
Question 11-6
We can’t square both sides of a “strictly smaller than” inequality and be sure that we’ll get another valid statement. we get 9 < 6. which is false.” and “logically implies. How would we write this fact entirely in logical symbols?
Answer 11-5
Remember the symbols for “strictly smaller than.
. or mathematical expressions p and q. and it remains the same if we multiply through by a positive real number.
Question 11-5
If a quantity x is strictly smaller than another quantity y. If p is strictly larger than q. If we square both sides.
Answer 11-6
Consider the fact that −3 < 2.Part Two 309 Answer 11-2
If we have two numbers.
Question 11-3
Suppose we multiply an inequality through by a nonzero real number. and it remains the same if we divide through by a positive real number. variables. we write p > q. we write p ≤ q. then five types inequalities can exist: • • • • • If p is strictly smaller than q. we write p < q. Under what circumstances will the sense of the inequality be reversed? Under what circumstances will the sense of the inequality stay the same?
Answer 11-3
The sense of the inequality is reversed if we multiply through by a negative real number.” “smaller than or equal to. Provide an example that shows why. If p is larger than or equal to q.

We see the following equation: x − 2a = 5a How can we get an equation with x alone on the left side of the equals sign and a multiple of a alone on the right?
Answer 12-1
We can morph the equation as follows: x − 2a = 5a (x − 2a) + 2a = 5a + 2a x = 7a
Question 12-2
Suppose b is a constant and z is a variable for which we want to solve. or we’ll run into trouble of some sort. it’s best to avoid dividing through by any expression that contains the variable. The fact that the expression contains an unknown means that we can’t be sure it’s nonzero until we know the solution to the equation! When trying to solve a first-degree equation. therefore.Part Two 311
Chapter 12
Question 12-1
Suppose a is a constant and x is a variable for which we want to solve. there’s a hidden risk.
. What is that risk?
Answer 12-3
The expression containing the variable must not be equal to 0. We see: z + 3b = b/2 How can we get an equation with z alone on the left side of the equals sign and a multiple of b alone on the right?
Answer 12-2
We can morph the equation as follows: z + 3b = b/2 2(z + 3b) = 2(b/2) 2z + 6b = b (2z + 6b) − 6b = b − 6b 2z = −5b z = (−5/2)b
Question 12-3
When we divide both sides of a first-degree equation by any expression containing the variable.

b. or N stand for a general constant if the subject happens to be chemistry. then we will find ourselves dividing by 0 on the right side of the equation. If c and d happen to be additive inverses. Why?
Answer 12-6
Some letters are widely used to represent specific physical. We don’t want to let c be a general constant if we’re writing about relativity theory. we must restrict the values of the constants. Letters x.
. c represents the speed of light in a vacuum. or e indicate a general constant if the discussion involves exponential functions. because it contains the square of the variable. N is often used to represent a chemical constant called the Avogadro constant.
Question 12-6
When using letters to represent constants.312 Review Questions and Answers Question 12-4
Suppose we see the following equation where a. and c represent constants. That is. we have to be clear and careful with the context.
Question 12-5
Which of the following equations are first-degree equations in one variable? Which are not? Letters a. and d are constants and x is the variable: 3abx = 7x /(c + d ) To be sure this equation makes sense. because they contain more than one variable. and z represent variables. b. x + 3a + 2b = c (c + a)x = b (x + y)a = b y2 + ay + b = c x+y+z=0
Answer 12-5
The first and second equations are first-degree equations in one variable. c. or mathematical constants. e is commonly used to represent the exponential constant. chemical. How?
Answer 12-4
We must require that c ≠ −d. In physics. y. The fourth equation isn’t. The third and fifth equations aren’t. we can’t let c and d be additive inverses. or when reading texts in which letters are used to represent constants.

showing and justifying every step in an S/R table? x − 2c + 7 = 4x + 2ax + b + c
Answer 12-7
We can morph the equation as described in Table 20-1.Part Two 313 Question 12-7
Suppose a. We solve for x in terms of the constants. which is addressed in Question and Answer 12-8. The end result is x = (b + 3c − 7) / (−3 − 2a)
Question 12-8
We had better be careful about something in the last step of the process shown by Table 20-1. and an expression containing the constants without x on the right side. How can we manipulate the following equation so it contains x all by itself on the left side. What’s that?
Answer 12-8
We must require that 2a ≠ −3.
Question 12-9
What is the standard form for a first-degree equation in one variable?
Table 20-1. and x is a variable. b. Equation morphing process for Answer 12-7.
Statements x − 2c + 7 = 4x + 2ax + b + c x + 7 = 4x + 2ax + b + 3c x = 4x + 2ax + b + 3c − 7 x − 4x = 2ax + b + 3c − 7 x − 4x − 2ax = b + 3c − 7 x + (−4x) + (−2a)x = b + 3c − 7 [1 + (−4) + (−2a)]x = b + 3c − 7 (−3 − 2a)x = b + 3c − 7 x = (b + 3c − 7)/(−3 − 2a) Reasons This is the equation we are given Add 2c to each side Subtract 7 from each side Subtract 4x from each side Subtract 2ax from each side Change subtractions to negative additions on left side of equation Right-hand distributive law on left side of equation Simplify left side of equation Divide each side by (−3 − 2a)
. we allow for the possibility of division by 0 in the last step. If we do not impose that restriction. But there’s a catch. and c are constants.

and D. then the standard form is ax + b = 0 Any first-degree equation in one variable can always be put into this form.314 Review Questions and Answers Answer 12-9
If a and b are constants and x is the variable. but this basic form can always be derived. as follows: (−3 − 2a) = p and −(b + 3c − 7) = q Now we have px + q = 0
Chapter 13
Question 13-1
Figure 20-1 shows a mapping between sets. into the standard form for a first-degree equation in one variable. The five points in set B represent all the elements in that set. c. they “tell the whole story. and numerals. C. this equation is in the standard form for a first-degree equation in one variable. But we might want to rename the expressions made up of a. and the constants a and b might be expressed in terms of other constants and numbers. and shown in the last line of Table 20-1. B. b. The dashed curves represent the entire mapping.” Furthermore. we can subtract the quantity (b + 3c − 7) from each side. Which set is the maximal domain? Which set is the co-domain? Which set is the essential domain? Which set is the range?
. B ⊂ A and C ⊂ D. and the five elements in set C represent all the elements of that set. Let’s use single letters p and q. Four sets are identified.
Answer 12-10
Here’s the equation in the form we got when we solved for x in terms of the constants: x = (b + 3c − 7)/(−3 − 2a) We can multiply each side by (−3 − 2a) to obtain (−3 − 2a)x = b + 3c − 7 Next.
Question 12-10
Put the equation derived in Answer 12-7. labeled A. The variable might be called something other than x. getting (−3 − 2a)x − (b + 3c − 7) = 0 Technically.

That is.Part Two 315
C A
B
D
Figure 20-1 Illustration for Questions and Answers 13-1
through 13-8. note that each of the points in set B maps to a single point in set C.
Question 13-3
In Fig 20-1. it is a surjection. what can we say about the mapping?
Answer 13-2
The mapping is onto set C.
Answer 13-1
The maximal domain is set A. The essential domain is set B. The range is set C. On this basis. what type of mapping is this?
. and vice-versa. suppose that each point in set B maps to a single point in set C.
Question 13-4
In Fig 20-1. That is. On that basis. it is an injection.
Question 13-2
In Fig 20-1. and according to what we know so far about the points and the sets. What can we say about the mapping on this basis?
Answer 13-3
The mapping is one-to-one. The co-domain is set D. the five points within set C represent the entire set.

b3 maps to c3. and on the descriptions given so far. b4 maps to c2. (b4. The only difference is that there is an extra element in set B.
Question 13-9
Figure 20-2 shows a mapping that’s almost the same as the mapping of Fig. How can we state this mapping as a set of ordered pairs?
Answer 13-7
We can state the mapping as the set {(b1. what type of relation is the mapping between B and C?
Answer 13-5
The mapping is a function.
Question 13-6
Suppose the sense of the mapping in Fig.c5). Would that inverse be a function?
Answer 13-6
Yes. (b3. because no single element in set B is mapped to more than one element in set C. 20-1 by the names b1 through b5. 20-1. which elements are values of the independent variable? Which elements are values of the dependent variable?
Answer 13-8
The elements b1 through b5 are values of the independent variable. because it is both a surjection and an injection.316 Review Questions and Answers Answer 13-4
The mapping is a bijection between B and C. and it maps to one of the existing elements
. (b5. and b5 maps to c1. 20-1.
Question 13-7
Let’s call the five points in set B of Fig. (b2.c3).c1)}
Question 13-8
In the ordered pairs given in Answer 13-7 relevant to Fig. Suppose that b1 maps to c5.c4). b2 maps to c4. so that it went from set C to set B. 20-1. This would give us the inverse of the relation from B to C. and the elements c1 through c5 are values of the dependent variable. 20-1 were reversed. and the five points in set C by the names c1 through c5. because no single element in set C would map to more than one element in set B.
Question 13-5
Based on Fig.c2).

because it’s not one-to-one. the dashed curves represent the entire mapping. This mapping is a relation. because it maps to every element of set C. As before. 20-2 as a relation. Now suppose the sense were reversed. because no single element in set B maps to more than one element in set C. Is this new mapping a surjection? Is it an injection? Is it a bijection? Is it a relation? Is it a function?
Answer 13-9
The mapping of Fig. It’s not an injection.Part Two 317
C A
B
D
Figure 20-2 Illustration for Questions and Answers 13-9
and 13-10. 20-2 is a surjection. and the five elements in set C represent all the elements of that set. The six points in set B represent all the elements in that set. How are those quadrants normally oriented? How are the positive and negative values of the variables portrayed in those quadrants?
. giving us the inverse of the relation.
Chapter 14
Question 14-1
The Cartesian plane is divided into four quadrants. however.
Question 13-10
Consider the mapping of Fig. so that the mapping went from set C to set B. because one of the elements of set C would map to two elements of set B.
in set C. That means it is not a bijection. It’s also a function. Would that inverse be a function?
Answer 13-10
No.

20-3.318 Review Questions and Answers Answer 14-1
The first quadrant of the Cartesian plane is usually the one at the upper right.−3). and is represented by the ordered pair (−5. called P. In the third quadrant. and R. Q. plotted in the Cartesian plane. it is represented by (0. and both variables are positive.
Question 14-3
Imagine horizontal lines (that is. Point R is in the first quadrant. In the fourth quadrant.4). Point Q is not in any quadrant because it is directly on one of the axes. the independent variable is negative and the dependent variable is positive.−1). and is represented by (2. which is usually at the lower right. lines parallel to the x axis) running through each of the three points in Fig. In which quadrants do these points lie? What are the ordered pairs representing these points?
Answer 14-2
Point P is in the third quadrant. the independent variable is positive and the dependent variable is negative. What are the equations of these lines? Do any of them represent functions of x ?
y 6 4 2 x –6 –4 –2 –2 P –4 –6 Q 2 4 6 R
Figure 20-3 Illustration for Questions and Answers
14-2 through 14-6. which is usually at the lower left. In the second quadrant. which is usually at the upper left.
. both variables are negative.
Question 14-2
Figure 20-3 shows three points.

Question 14-6
Suppose both of the coordinates of each of the points in Fig. because they all fail the vertical-line test. 1). In which quadrants will these new points appear?
Answer 14-5
The coordinates of P* will be (−3. 20-3 are multiplied by −1. The coordinates of Q* will be (−1. just as is P. a line parallel to the y axis) will never intersect any one of those three horizontal lines at more than one point. What are the equations of these lines? Do any of them represent functions of x ?
Answer 14-4
A vertical line through point P has the equation x = −5. All three of these lines represent functions of x. A vertical line through point R has the equation x = 2. 20-3. −5). that is. so it will be in the third quadrant. A horizontal line running through point Q has the equation y = −1.Part Two 319 Answer 14-3
A horizontal line through point P has the equation y = −3. 2). Call the new points P*. The coordinates of −Q will be (0. −4). so P* will be in the third quadrant.
Question 14-5
Suppose the coordinates of each of the points in Fig. 20-3 are transposed. because they all pass the vertical-line test. A movable vertical line that intersects any of those three lines at one point will intersect it at infinitely many other points as well. None of these lines represent functions of x. so −P will be in the first quadrant. neither curve F nor curve H are functions of x. The coordinates of R* will be (4. because they both fail the vertical-line test. A vertical line running through point Q has the equation x = 0. so it will not be in any quadrant.
. A movable vertical line (that is. the order of each ordered pair is reversed. The coordinates of −R will be (−2. In which quadrants will these new points appear?
Answer 14-6
The coordinates of −P will be (5. but on the positive y axis. 0). so it will be in the first quadrant. we can see that curve E and line G both represent functions of x. and −R.
Question 14-4
Imagine vertical lines running through each of the three points in Fig. and R*. so it will not be in any quadrant. just as is R. 3). −Q. Which of these relations are functions of x ? How can we tell?
Answer 14-7
Using the vertical-line test.
Question 14-7
Figure 20-4 shows the graphs of four different relations between x and y in Cartesian coordinates. A horizontal line through point R has the equation y = 4. Call the new points −P. They both pass the test. Q*. However. at least within the viewing region of this graph. but on the negative x axis.

we can bring the two vertical lines together so they both intersect the curve at the extreme left point. but those are trivial functions! In the case of curve H.” we always get two values of y for each value of x when we do this with either F or H. so we have mappings from values of y to values of x. that’s a trivial result. That restricts the domain so severely that we get mappings of one point onto one other point. thereby restricting the domain. A convenient way to imagine these inverses is to let y be the independent variable and let x be the dependent variable. Again. Which of the inverses. 20-4. how?
Answer 14-8
Imagine “taking slices” of curve F or curve H by considering only that portion of the curve that falls between two movable vertical lines. The only way we can get a function out of curve F is to bring the two vertical lines together so they both intersect F at the extreme left point or the extreme right point of the ellipse. No matter how we “slice it. defined in this way. This limits the values of x that apply to the curves.
Question 14-9
Consider the inverses of the relations shown in Fig.
Question 14-8
Can we restrict the domains of relations shown by curves F or H to make either of them represent a nontrivial function of x ? If so.320 Review Questions and Answers y E 6 G 4 2 x –6 F –4 –6 H –4 –2 –2 2 4 6
Figure 20-4 Illustration for Questions and Answers 14-7
through 14-10. are functions?
.

QR.)
Question 14-10
Can we restrict the domains of the inverses of relations E or F to make either of them into a nontrivial function of y? If so. The only way we can get a function out of the inverse of curve F is to bring the two horizontal lines together so they both intersect the graph at the extreme top point or the extreme bottom point. each extending indefinitely in either direction. They both pass the test.−3) and Q = (0.” we always get two values of x for each value of y. because they fail the horizontal-line test. we can see that G and H represent functions of y. how?
Answer 14-10
In the case of the inverse of the “bent line” curve E. Let’s call the lines PQ. (It’s visually apparent which is which!) Assume that the ordered pairs for the points are all pairs of integers. This is based on the assumption that the lines that make up curve E continue straight for infinite distances in both directions. and R as we saw in Fig. Based on this information. thereby restricting the domain. assume that the points are exactly where they appear to be.
Chapter 15
Question 15-1
Figure 20-5 is a Cartesian graph showing the same three points P. Those are trivial results. In other words. (Part of curve E is itself a horizontal line. Q. But neither E nor F represents functions of y. No matter how we “slice it. The inverse of curve F is not so cooperative. This limits the values of y that apply to the curve. Δy = −1 − (−3) = −1 + 3 =2 and Δx = 0 − (−5) =0+5 =5
. 20-3. how can we determine the slope of line PQ? How can we determine the y-intercept of line PQ?
Answer 15-1
The slope of a line is equal to the change in the y-value (Δy) divided by the change in the x-value (Δx) as we move from one point on the line to another point. and PR. Imagine “taking slices” of by considering only that portion of the curve that falls between two movable horizontal lines. pass through pairs of these points. we can make it into a function if we confine the domain to values of y strictly larger than 1. Therefore.−1). We know the ordered pairs for the two points as P = (−5. Three lines. at least within the viewing region.Part Two 321 Answer 14-9
Using the horizontal-line test.

As with line PQ. which is the slope of the line.322 Review Questions and Answers y 6 4 2 x –6 –4 –2 –2 P –4 –6 Q 2 4 6 R
Figure 20-5 Illustration for Questions and Answers
15-1 through 15-10. which is the slope of the line.
That means Δy /Δx = 2/5. That value is −1. because Q happens to lie on the y axis. Therefore.4).−1) and R = (2. 20-5. which is the y-value at point Q. how can we determine the slope of line QR ? The y-intercept?
Answer 15-2
We know the ordered pairs for the two points as Q = (0. Δy = 4 − (−1) =4+1 =5 and Δx = 2 − 0 =2 That means Δy /Δx = 5/2.
Question 15-2
In Fig. the y-intercept is −1.
. The y-intercept is the y-value at Q.

Δy = 4 − (−3) =4+3 =7 and Δx = 2 − (−5) =2+5 =7 That means Δy /Δx = 7/7 = 1. Remember the general slope-intercept form for a line in Cartesian coordinates: y = mx + b where x is the independent variable. and PR ?
Answer 15-4
Now that we know the slopes and the y-intercepts of all three lines. If we increase the x-value of point P by 5 units. we arrive at the y axis.Part Two 323 Question 15-3
In Fig. which is the slope of the line. For line PQ. we can write the slopeintercept equations straightaway. so 5 more than that is 2. m is the slope. and b is the y-intercept. and 15-3. Note that if we move to the right from point P by Δx units. 15-2. QR. and we’ll be at a point 5 units above the y-value of P. we have y=x+2
.−3) and R = (2. what are the slope-intercept forms of the equations for lines PQ. The y-intercept of line PR is therefore equal to 2. we have y = (5/2)x − 1 For line PR. how can we determine the slope of line PR ? The y-intercept?
Answer 15-3
We know the ordered pairs for the two points as P = (−5. Therefore. The y-value of P is −3. The y-intercept can be inferred.4). we have y = (2/5)x − 1 For line QR.
Question 15-4
Based on Answers 15-1. y is the dependent variable. 20-5. we must go up by the same number of units to stay on the line.

We also know that for line PQ. based on the coordinates of point R and the slope of the line?
Answer 15-6
We know that R = (2. We also know that for line PR. so x0 = −5 and y0 = −3. so x0 = −5 and y0 = −3. y is the dependent variable. so x0 = 2 and y0 = 4. We know that P = (−5. the point-slope equation for line PQ is y − (−3) = (2/5)[x − (−5)] which can be simplified to y + 3 = (2/5)(x + 5)
Question 15-6
How can we determine the point-slope form of the equation for line QR. −3). Therefore. based on the coordinates of point P and the slope of the line?
Answer 15-5
Remember the point-slope form for a straight line in Cartesian coordinates: y − y0 = m(x − x0) where x is the independent variable. the slope m is equal to 2/5. the slope m is equal to 5/2. We also know that for line QR. and m is the slope of the line. the point-slope equation for line QR is y − 4 = (5/2)(x − 2)
Question 15-7
How can we determine the point-slope form of the equation for line PR.324 Review Questions and Answers Question 15-5
How can we determine the point-slope form of the equation for line PQ. the slope m is equal to 1. (x0. y0) are the coordinates of a point on the line. based on the coordinates of point P and the slope of the line?
Answer 15-7
We know that P = (−5. Therefore. 4). Therefore. the point-slope equation for line PR is y − (−3) = x − (−5) which can be simplified to y+3=x+5
. −3).

getting y−4=x−2 That’s all there is to it!
Question 15-10
Starting with the slope-intercept forms. how can we morph the equations for lines PQ. so x0 = 2 and y0 = 4. QR. How can we prove it by showing that the equations are equivalent?
Answer 15-9
If we can convert one of the equations into the other using the rules for equation morphing. the slope m is equal to 1.Part Two 325 Question 15-8
How can we determine the point-slope form of the equation for line PR. Let’s start with y+3=x+5 We can subtract 7 from each side. and PR in Fig. and c are integer constants?
Answer 15-10
From Answer 15-4. it will prove that the equations are equivalent. b. based on the coordinates of point R and the slope of the line?
Answer 15-8
We know that R = (2. the point-slope equation for line PR is y−4=x−2
Question 15-9
It’s intuitively obvious that the equations we derived in Answers 15-7 and 15-8 must represent the same line. 20-5 into the form ax + by = c where a. We also know that for line PR. the slope-intercept form of the equation for line PQ is y = (2/5)x − 1 We can multiply through by 5 to obtain 5y = 2x − 5 Subtracting 2x from each side gives us −2x + 5y = −5
. Therefore. 4).

. first for one variable and then for the other. the slope-intercept form of the equation for line QR is y = (5/2)x − 1 When we multiply each side by 2. We proceed like this: • Morph both equations into SI form with y all by itself on the left side of the equals sign. without substituting either variable for the other?
Answer 16-1
We can go through the morph-and-mix process twice.326 Review Questions and Answers
That’s in the form we want! Again from Answer 15-4. How can we solve such a system by morphing and mixing alone. we get 2y = 5x − 2 Subtracting 5x from each side. we learned how a two-by-two linear system in variables x and y can be solved by the following process: • Morph both equations into SI form with y all by itself on the left side of the equals sign. • Solve the first-degree equation for x. we obtain −5x + 2y = −2 That’s in the form we want! Once again referring to Answer 15-4. • Mix the two equations to get a first-degree equation in x. 16. • Mix the two equations to get a first-degree equation in x. • Substitute that solution back into one of the SI equations to solve for y. the slope-intercept form of the equation for line PR is y=x+2 Subtracting x from each side gives us −x + y = 2 That’s in the form we want!
Chapter 16
Question 16-1
In Chap.

Here it goes. we can do this: x − 3y + 9 = 0 −3y + 9 = −x −3y = −x − 9 3y = x + 9 y = (1/3)x + 3
Question 16-3
How can we combine the two equations from Answer 16-2 to get a first-degree equation and solve the original system for x?
Answer 16-3
We can mix the right sides of the two SI equations together and then solve the resulting firstdegree equation in x by manipulation. • Morph both equations into SI form with x all by itself on the left side of the equals sign. one step at a time:
.
Question 16-2
How can we put the following two-by-two linear system into a pair of SI equations with y all by itself on the left side of the equals sign? 2x − y + 8 = 0 and x − 3y + 9 = 0
Answer 16-2
By now. we’re good enough at equation manipulation to write down the steps one after another. • Mix the two equations to get a first-degree equation in y. For the first original equation. we can do this: 2x − y + 8 = 0 −y + 8 = −2x −y = −2x − 8 y = 2x + 8 and for the second original equation.Part Two 327
• Solve the first-degree equation for x. • Solve the first-degree equation for y. without having to justify everything.

328 Review Questions and Answers
2x + 8 = (1/3)x + 3 6x + 24 = x + 9 6x + 15 = x 5x + 15 = 0 5x = −15 x = −3
Question 16-4
How can we put the two-by-two linear system from Question 16-2 into a pair of SI equations with x all by itself on the left sides of the equals signs?
Answer 16-4
For the first original equation. as follows: (1/2)y − 4 = 3y − 9 y − 8 = 6y − 18 y + 10 = 6y 10 = 5y y=2
Question 16-6
How can we be sure the solution we obtained in Answers 16-3 and 16-5 is in fact the correct solution to the original two-by-two linear system?
. we can do this: x − 3y + 9 = 0 x + 9 = 3y x = 3y − 9
Question 16-5
How can we combine the two equations from Answer 16-4 to get a first-degree equation and solve the original system for y?
Answer 16-5
We can mix the right sides of the two SI equations together and then solve the resulting firstdegree equation in y by manipulation. we can do this: 2x − y + 8 = 0 2x + 8 = y 2x = y − 8 x = (1/2)y − 4 and for the second original equation.

Question 16-7
Do you suspect that I concocted the above problem so it would come out with a pair of “clean integers” for the solution? If so. we proceed like this: 2x − y + 8 = 0 2 × (−3) − 2 + 8 = 0 −6 − 2 + 8 = 0 0=0 It checks out! For the second original equation. we do this: x − 3y + 9 = 0 −3 − (3 × 2) + 9 = 0 −3 − 6 + 9 = 0 0=0 It checks again! Now we know the solution we obtained is correct. getting the sum −6x + 3y − 24 = 0 x − 3y + 9 = 0 −5x − 15 = 0
. Finally. using the solution point as the reference for both lines.Part Two 329 Answer 16-6
The solution we have obtained is x = −3 and y = 2. we can convert the point-slope equations to some other form to get the test problem. For extra credit. For the first original equation. Then we can write down the equations in point-slope form. here are the equations again: 2x − y + 8 = 0 x − 3y + 9 = 0
Answer 16-8
We can multiply the first equation through by −3 and then add it to the second equation. We must plug these values into both of the original equations to be certain we’ve gotten the right solution. you can try this and then solve the test problem you’ve created. and assign different slopes to those lines.
Question 16-8
How can we add multiples of the two original equations stated in Question 16-2 to solve the linear system for x ? For reference. you are right! How can we compose a two-by-two linear system as a test problem (for someone else to solve). and be sure the solution will turn out to be a pair of integers?
Answer 16-7
We can choose a point where the graphs of two lines intersect.

we add 15 to each side and then divide through by −5. To manipulate the equation. like this: 5y − 10 = 0 5y = 10 y=2
Question 16-10
How can we solve the two-by-two linear system stated in Question 16-2 by the rename-andreplace (substitution) method? Here are the original equations again. as follows: −5x − 15 = 0 −5x = 15 x = −3
Question 16-9
How can we add multiples of the two original equations stated in Question 16-2 to solve the linear system for y ?
Answer 16-9
We can multiply the second equation through by −2 and then add it to the first equation. we add 10 to each side and then divide through by 5.330 Review Questions and Answers
To solve this. getting the sum 2x − y + 8 = 0 −2x + 6y − 18 = 0 5y − 10 = 0 To solve this. so one of the variables appears all alone on the left side of the equals sign. we proceed just as we did in Answer 16-2: 2x − y + 8 = 0 −y + 8 = −2x −y = −2x − 8 y = 2x + 8
. for reference: 2x − y + 8 = 0 and x − 3y + 9 = 0
Answer 16-10
We start by converting either of the two equations to SI form. Let’s use the first equation and isolate y on the left side.

Then we proceed as follows: 2x − y + 8 = 0 2 × (−3) − y + 8 = 0 −6 − y + 8 = 0 −6 + 8 = y y=2
Chapter 17
Question 17-1
Let’s consider again the two-by-two system we saw in Question 16-2. How can we graph this system in Cartesian coordinates. Let’s use the first equation. we substitute the quantity (2x + 8) for y in the second equation and solve the result for x. and the solution of the system to find a third point that lies on both lines. as follows: x − 3y + 9 = 0 x − 3(2x + 8) + 9 = 0 x − 6x − 24 + 9 = 0 −5x − 15 = 0 −5x = 15 x = −3 Now that we know the value of x. they are: y = 2x + 8
. with x as the independent variable and y as the dependent variable? Here are the original equations: 2x − y + 8 = 0 and x − 3y + 9 = 0
Answer 17-1
We can use the SI forms of the equations to find their y-intercepts. we can plug it into either equation and solve for y. We’re lucky here. The SI forms of the equations were derived in Answer 16-2. because the intersection point is fairly far away from the y axis. Respectively.Part Two 331
Next.

3) is on the second line. 8). In the opposite sense.8)
y = (1/3)x + 3 Solution = (–3.2) (0.
.3)
x Each axis increment is 1 unit
y = 2x + 8
Figure 20-6 Illustration for Questions and Answers
17-1 through 17-6. which is the y-intercept. Figure 20-6 shows plots of these three points. 2). but the purpose of this exercise is to demonstrate the geometric principles of slope and intercept. If we move in the negative y direction
y
(0.332 Review Questions and Answers
and y = (1/3)x + 3 We know from these equations that the y-intercepts are 8 and 3. 20-6? (We can set y = 0 and then calculate x by arithmetic.)
Answer 17-2
The slope of the line is 2. That point lies on both lines. as we’ve already determined. Let’s start at the point (0. if we start from any point on the line and move in the negative y direction by p units. we must move in the positive y direction by 2n units to stay on the line.
Question 17-2
How can we determine the x-intercept of the line representing the equation y = 2x + 8 on the basis of the known slopes and the point data in Fig. Therefore. The solution to the system. so the point (0. we must move in the negative x direction by p /2 units to stay on the line. 8) is on the first line and the point (0. if we start from any point on the line and move in the positive x direction by n units. along with plots of the lines connecting the points and representing the equations. is (−3.

but again. and manipulate things until we get the SI form with x all alone on the
. now we know it’s really true.Part Two 333
along the line until y = 0. It looks that way in the figure.)
Answer 17-4
The slope of the line is 1/3. that was done starting with the original equation. or 4. 20-6.
Question 17-3
How can we verify that the preceding answer is correct by manipulating the equation for the line?
Answer 17-3
We can start with either the original equation or the SI form with y as the dependent variable. In the opposite sense.
Question 17-4
How can we determine the x-intercept of the line representing the equation y = (1/3)x + 3 on the basis of the known slopes and the point data in Fig. if we start from any point on the line and move in the positive x direction by n units.
Question 17-5
How can we verify that the preceding answer is correct by manipulating the equation for the line?
Answer 17-5
As we did in Answer 17-3. we must move in the positive y direction by n /3 units to stay on the line. and we’ll have gone 3 × 3. If we move in the negative y direction along the line until y = 0. That means the x-intercept is −4. and manipulate things until we get the SI form with x all alone on the left side of the equals sign. we can proceed like this: y = 2x + 8 y − 8 = 2x (1/2)y − 4 = x x = (1/2)y − 4 This SI equation tells us that the x-intercept is equal to −4. 3). That means the x-intercept is −9. we’ll end up on the x axis. if we start from any point on the line and move in the negative y direction by p units. we must move in the negative x direction by 3p units to stay on the line. we can start with either the original equation or the SI form with y as the dependent variable. Therefore. we’ll end up on the x axis. units in the negative x direction from the y axis. we can set y = 0 and then solve for x. In Answer 16-4. 20-6? (As in Question 17-2. If we start with the SI form with y as the dependent variable. Let’s start at the point (0. This is outside the field of view in Fig. units in the negative x direction from the y axis. which is the y-intercept. this exercise is meant to show how slope and intercept are related geometrically. or 9. and we’ll have gone 8/2.

20-6.
Question 17-6
How can we morph Fig.–3)
Figure 20-7 Illustration for Questions and Answers
17-6 through 17-10. we can do this: y = (1/3)x + 3 y − 3 = (1/3)x 3y − 9 = x x = 3y − 9 This SI equation tells us that the x-intercept is equal to −9. that was done starting with the original equation. 20-6 and rotate the entire assembly as a single mass—axes. If we start with the SI form with y as the dependent variable. to get a graph that will show the same system with y as the independent variable and x as the dependent variable?
Answer 17-6
We can use the rotate-and-mirror method. lines.
. the graph of the system from Question and Answer 17-1 and Fig. 20-7. Then we mirror the whole thing left-to-right. and points—counterclockwise by a quarter-turn (90°). We start with Fig. Finally.334 Review Questions and Answers
left side of the equals sign.0)
(8. The result is shown in Fig.
Question 17-7
What are the SI forms of the equations for the lines in Fig. 20-7?
x
Each axis increment is 1 unit
(3. 20-6.0) y
Solution = (2. we reverse the ordered pairs to obtain the new points. In Answer 16-4.

0). we must go 4 × 3. parallel to the x axis and running through the point (−2. That will get us to the point (−2. Now let’s move along the line for the equation x = 3y − 9. units in the negative x direction to stay on the line. It’s −4. −15). The slope is 1/2. The slope is 3. we can travel in the negative y direction by 4 units along either of our existing lines. That means if we go 4 units in the negative y direction. at what points would the lines representing the two-by-two linear system intersect the graph of the equation y = −2?
Answer 17-10
The graph of the equation y = −2 would appear as a vertical line in Fig. To reach this line from the point (2. Therefore.
Question 17-9
What is the x-intercept of the line representing the equation x = 3y − 9?
Answer 17-9
Again. and on the point data shown in Fig. First.
Question 17-10
Based the known slopes of the lines. 20-7.
Chapter 18
Question 18-1
Here is a set of equations that forms a three-by-three linear system: 4x = 8 + 4y + 4z 2y = 5 + x − 5z 4z = 13 − 2x + y
.Part Two 335 Answer 17-7
They are the equations we derived in Answers 17-3 and 17-5: x = (1/2)y − 4 and x = 3y − 9
Question 17-8
What is the x-intercept of the line representing the equation x = (1/2)y − 4?
Answer 17-8
We can answer this straightaway. units in the negative x direction to stay on the line. −5). because the equation is in SI form with x as the dependent variable. −3). we can infer this from the SI equation having x as the dependent variable. 20-7. if we go 4 units in the negative y direction. let’s move along the line for x = (1/2)y − 4. or 12. That will put us at the point (−2. we must go 4/2. or 2. It’s −9.

starting with the original equation: 2y = 5 + x − 5z −x + 2y = 5 − 5z −x + 2y + 5z = 5
Question 18-3
How can we put the third equation in Question 18-1 into the form ax + by + c = d. c. starting with the original equation: 4z = 13 − 2x + y 2x + 4z = 13 + y 2x − y + 4z = 13
Question 18-4
Based on the rearrangements in Answers 18-1 through 18-3. like this: 4x − 4y − 4z = 8 −x + 2y + 5z = 5 2x − y + 4z = 13
. b. b. c. and d are constants?
Answer 18-3
Here are the steps we can take. and d are constants?
Answer 18-1
Here are the steps we can take. one at a time. c. starting with the original equation: 4x = 8 + 4y + 4z 4x − 4y = 8 + 4z 4x − 4y − 4z = 8
Question 18-2
How can we put the second equation in Question 18-1 into the form ax + by + c = d.336 Review Questions and Answers
How can we put the first of these equations into the form ax + by + c = d. b. where a. one at a time. how can we state the three-bythree linear system from Question 18-1 now? What strategies can we use to solve it?
Answer 18-4
We can state the system by combining the final equations from Answers 18-1 through 18-3. and d are constants?
Answer 18-2
Here are the steps we can take. one at a time. where a. where a.

We can get rid of any of the three variables by morphing and adding any two pairs of the three-variable equations. resulting in a two-by-two linear system. getting −x + 2y + 5z = 5 4x − 2y + 8z = 26 3x + 13z = 31 Now we have the following two-by-two linear system in the variables x and z : x + 3z = 9 3x + 13z = 31
Question 18-6
How can we solve the above two-by-two system for z using the addition method?
. and then substituting back to solve for the variable we eliminated. let’s look at the second two equations from the three-by-three system as stated in Answer 18-4: −x + 2y + 5z = 5 2x − y + 4z = 13 We can multiply the bottom equation through by 2 and then add it to the top equation. That’s the method we’ll use here. 18.Part Two 337
We can solve this system in many different ways. using the first two equations and then the second two?
Answer 18-5
Here are the first two equations from the three-by-three linear system as stated in Answer 18-4: 4x − 4y − 4z = 8 −x + 2y + 5z = 5 We can divide the top equation through by 2 and then add the bottom equation. To get the second equation.
Question 18-5
How can we obtain a two-by-two linear system in x and z from the three-by-three system as stated in Answer 18-4. we learned to solve systems of this type by getting rid of one variable. solving that system. getting the sum 2x − 2y − 2z = 4 −x + 2y + 5z = 5 x + 3z = 9 That’s the first equation in our two-by-two system. In Chap.

how can we find the value of y?
Answer 18-8
We can plug in the value 6 for x and the value 1 for z in any of the original three equations. but still tentative.
Question 18-8
Now that we know the values of x and z in the original three-by-three system. We have x + 3z = 9 x+3×1=9 x+3=9 x=6 Now we have the tentative solutions x = 6 and z = 1. Let’s use the top equation. solution to the original three-by-three system: x=6 y=3 z=1
. This gives us the sum −3x − 9z = −27 3x + 13z = 31 4z = 4 This simplifies to the tentative solution z = 1. Let’s use the first one: 4x = 8 + 4y + 4z 4 × 6 = 8 + 4y + 4 × 1 24 = 8 + 4y + 4 20 = 8 + 4y 12 = 4y 3=y Now we have the complete. as they are stated in Question 18-1.
Question 18-7
How can we solve the above two-by-two system for x by substituting in the solution for z we obtained in Answer 18-6?
Answer 18-7
We can plug in the value 1 for z in either of the equations in the two-by-two system as stated at the end of Answer 18-5.338 Review Questions and Answers Answer 18-6
Let’s multiply the top equation through by −3 and then add the bottom equation to it.

we proceed directly from left to right after the equals sign. so the whole sequence goes like this:
. We did it indirectly for the first equation in Answer 18-8. but if we want to be completely rigorous. we must plug all three values into that equation along with the other two. we get 4x = 8 + 4y + 4z 4×6=8+4×3+4×1 24 = 8 + 12 + 4 24 = 24 Check one. and then we add 3. we get an extra step. (What if we made a mistake in Answer 18-8? Don’t laugh. for reference: 4x = 8 + 4y + 4z 2y = 5 + x − 5z 4z = 13 − 2x + y Grinding out the numbers in the first equation. In the second equation. If we do that in the above calculation. we can change the subtraction to negative addition. We subtract 12.Part Two 339 Question 18-9
How can we be sure that the solution we have obtained for the three-by-three system presented in Question 18-1 is correct?
Answer 18-9
We must substitute our solutions into all three of the original equations. In the third equation. We don’t subtract the quantity (12 + 3)! If we have any doubts when we come across a situation like this. we get 2y = 5 + x − 5z 2×3=5+6−5×1 6=5+6−5 6=6 Check two. Things like that can and do happen!) Here are the original three equations again. we get 4z = 13 − 2x + y 4 × 1 = 13 − 2 × 6 + 3 4 = 13 − 12 + 3 4=4 Check three. Mission accomplished! Note that in the mixed addition/subtraction here.

and place the remaining numerals neatly in the cells of a table. and the third equation is represented by the bottom row. and/or use different equations to get the two-by-two linear system in the intermediate phase. we could eliminate a different variable in the beginning. it means we’ve made a mistake somewhere. We could also start all over and approach the problem in another way. remove the variables.340 Review Questions and Answers
4z = 13 − 2x + y 4 × 1 = 13 − 2 × 6 + 3 4 = 13 − 12 + 3 4 = 13 + (−12) + 3 4=4
Question 18-10
Suppose that when we checked our solutions in the preceding answer.
Chapter 19
Question 19-1
Here is the three-by-three linear system taken from Answer 18-4: 4x − 4y − 4z = 8 −x + 2y + 5z = 5 2x − y + 4z = 13 How can we arrange these equations into a matrix that can be manipulated to solve this system?
Answer 19-1
We take the coefficients. The result looks like this:
4 −1 2 Question 19-2 −4 2 −1 −4 5 4 8 5 13
What general procedure can we use to solve the three-by-three system. After that. starting with this matrix?
. What could we do then?
Answer 18-10
Whenever we check solutions and discover that they don’t work out. we found that they did not work out. of course. For example. we’d have to check our solutions again for correctness. In that case. the second equation is represented by the middle row. we must go back through each step in our solution process and find the mistake. The first equation is represented by the top row of the matrix.

which looks like this:
1 0 0 0 1 0 0 0 1 x y z
Once we have the matrix in the unit diagonal form. assuming we haven’t made any mistakes. and then replace the elements in either row by the sum. if they are fractions. to their lowest forms. and an at sign (@) can represent any nonzero real number. and z represent the solution to the original three-by-three linear system.Part Two 341 Answer 19-2
We can play the matrix morphing game to get the matrix into echelon form. keeping the elements of both rows in the same order from left to right. which looks like this:
# 0 0 # # 0 # # # # # #
where a pound sign (#) can represent any real number. Then we continue the game and go for the diagonal form. y. which looks like this:
@ 0 0 0 @ 0 0 0 @ # # #
where a pound sign can represent any real number. • Multiply or divide all the elements in any row by a nonzero constant. • Add all the elements in any row to all the elements in another row. We can reduce the values we have found. we should plug our answers into the original equations to be sure we’ve arrived at the correct solution to the system. Finally. Then we must get the matrix into the unit diagonal form.
.
Question 19-3
How is the matrix morphing game played?
Answer 19-3
There are three types of moves: • Swap two rows. while keeping the elements of both rows in the same order from left to right. the values x. keeping the elements in the same order from left to right.

and then reduce the sizes of the elements to make the matrix easier to work with later?
Answer 19-4
Here’s the matrix again. we get
2 −2 0 −2 4 3 −2 10 14 4 10 23
Adding the first two rows and then replacing the second row with the sum. we obtain
2 0 0 −2 2 3 −2 8 14 4 14 23
We can multiply the second row by −3 and the third row by 2 to get
2 0 0 −2 −6 6 −2 −24 28 4 −42 46
.342 Review Questions and Answers Question 19-4
Using the rules outlined above. for reference:
4 −1 2 −4 2 −1 −4 5 4 8 5 13
We can multiply the second row by 2 to get
4 −2 2 −4 4 −1 −4 10 4 8 10 13
Now we can add the second and third rows and then replace the third row with the sum. how can we get the matrix from Answer 19-1 into echelon form. obtaining
4 −2 0 −4 4 3 −4 10 14 8 10 23
If we divide the first row by 2.

and the third row by 4. making it easier to work with as we continue the game. getting
1 0 0 −1 −1 0 −1 −4 1 2 −7 1
Question 19-5
Using the rules of the matrix morphing game. we get
2 0 0 −2 −6 0 −2 −24 4 4 −42 4
This matrix is in echelon form.Part Two 343
Adding the second row to the third and then replacing the third row with the sum. we get
1 0 0 0 1 0 3 4 −3 9 7 −3
. We can reduce the sizes of the numbers in this matrix. how can we get the last matrix in Answer 19-4 into diagonal form?
Answer 19-5
Multiplying the second row of the echelon-form matrix we just derived by −1. we get
1 0 0 −1 1 0 −1 4 1 2 7 1
Adding the first row to the second. we obtain
1 0 0 0 1 0 3 4 1 9 7 1
If we multiply the third row by −3. and then replacing the first row with the sum. Let’s divide the first row by 2. the second row by 6.

y. We get
1 0 0 0 1 0 0 4 −4 6 7 −4
Adding the second and third rows and then replacing the second row with the sum. and z in order: x=6 y=3 z=1 We know these values are correct. because we reduced our numbers along the way.
. We can divide the third row of the last matrix in Answer 19-5 by −4 to get
1 0 0 Question 19-7 0 1 0 0 0 1 6 3 1
What is the solution to the original three-by-three system based on Answer 19-6?
Answer 19-7
The solution can be read down the last column for x.344 Review Questions and Answers
Adding the first and third rows and then replacing the first row with the sum gives us
1 0 0 0 1 0 0 4 −3 6 7 −3
Now let’s multiply the third row by 4/3. because they are the same values we found and verified in the previous section for the same original three-by-three linear system. we obtain the diagonal matrix
1 0 0 Question 19-6 0 1 0 0 0 −4 6 3 −4
How can we get the last matrix in Answer 19-5 into unit diagonal form?
Answer 19-6
This is easy.

Part Two 345 Question 19-8
Suppose that we play the matrix morphing game in an attempt to solve a three-by-three linear system. we suspect we made a mistake somewhere. and we come up with a matrix that looks like this:
0 1 0 0 1 0 0 1 0 −5 2 7
At first. There are infinitely many solutions. What is the reason for this absurd result?
Answer 19-8
The original system of equations is inconsistent. It’s more like slow torture! But there’s some more good news: Computers can be programmed to play the matrix morphing game quickly and without pain.
. and so on?
Answer 19-10
Yes. There’s bad news. too: The number of steps in the process increases much faster than the dimension. matrix morphing is no game. That’s the good news. But when we try the game again. If carried out “manually” on a large system. There is no unique solution. five-byfive. The matrix morphing game can be applied to linear systems of any finite dimension. even for gigantic linear systems. such as four-by-four. and we come up with this elegant but useless matrix:
1 1 1 1 1 1 1 1 1 1 1 1
What does this tell us?
Answer 19-9
The original system of equations is redundant.
Question 19-9
Suppose that we play the matrix morphing game in an attempt to solve another three-by-three linear system.
Question 19-10
Can the matrix morphing game be used to solve larger systems. we get the same matrix.

Based on that fact. From this we can calculate j3 = j2 × j = −1 × j = −j Now for the 4th power: j4 = j3 × j = −j × j = −1 × j × j = −1 × j 2 = −1 × (−1) =1 And the 5th power: j5 = j4 × j =1×j =j
.350 Imaginary and Complex Numbers
you’re more likely to go into science or engineering than pure mathematics. Recall from Chap. Does the idea escape your “mind’s eye”? If so.
Positive and negative j The square root of −1 can have either of two values. just as can the square root of any positive real number. If we leave them out. These two numbers are not the same.
Solution
Keep in mind that j is the positive square root of −1. don’t worry about it.
Here’s a challenge!
All of the laws of real-number arithmetic also apply to the unit imaginary number. One of these is j. The parentheses are important in this expression. figure out what happens as j is raised to increasing integer powers starting with the 1st power. we can be sure that j 1 = j. which is equal to −1. just as the positive and negative square roots of 1 are not the same!
Are you confused?
Some people have trouble envisioning this unit imaginary number. so j isn’t any more bizarre than 0. so you should get used to the notation they prefer. 3 that the natural numbers—the simplest ones—are built up from a set containing nothing! All numbers are abstract in the literal sense. which is (−1)1/2. someone might get the idea that we’re discussing the quantity −(11/2). The other is −j. also called the j operator. the product of j and −1. Because all the laws of the reals also apply to j. By definition. or −1. or any other number. j 2 = −1.

The Imaginary Number Line
And the 6th: j6 = j5 × j =j×j = j2 = −1
351
Can you see what will happen if we keep going like this.
Be careful!
In the “challenge” calculation at the end of the previous section. If we want to multiply j by a negative real number −b. or a power of j (as in j 4). Engineers usually write the j before the real number. if n is a positive integer.” you might ask. which we would call j 2. j was raised to integer powers. keep in mind that they all refer to the same quantity. We must pay close attention to whether that real number is meant to be a multiple of j (as in j4). If we’re not careful. increasing the integer power by 1 over and over? We’ll go in a four-way cycle. we write jb. j 5 = (−1)1/2 × 251/2 = (−1 × 5)1/2 = (−5)1/2 and (−4)1/2 = (−1 × 4)1/2 = (−1)1/2 × 41/2 = j2 If we take the real number line and multiply the value of every point by j. If you see other notations for imaginary numbers such as 2j. In general. For example. 2i (the way most pure mathematicians write it). Conversely. If you grind things out.
. and so on. or even i 2. j n = j n+4
The Imaginary Number Line
The unit imaginary number j can be multiplied by any real number to get the positive square root of some negative real number. we write −jb.
Are you confused?
“Why. j 8 = 1. j 9 = j. the positive square root of any negative real number is equal to some positive-real multiple of j. If we want to multiply j by a positive real number b. you’ll see for yourself that j 7 = −j. putting the minus sign in front of j rather than between j and b. “do we write j before the real-number numeral and not after it?” It’s a matter of preference. j 10 = −1. we can confuse expressions like these with integer multiples of j. the result is the imaginary number line (Fig. 21-1).

In Fig. If you go upward on the line. The imaginary values are
defined according to the values of the real-number multiples of j.
Relative and absolute values Imagine a “number reflector” plane.
|j 3| = 3 and |−j 3| = 3
. perpendicular to the imaginary number line and passing through the point for j0. (Zero is the only real number that’s also imaginary. 21-1. like a mathematical mirror.) The definitions of the positive and negative imaginary numbers.” are analogous to the definitions for the real numbers. the distance of an imaginary number from the point for j0 is defined as its absolute value. The absolute value of an imaginary number is equal to the nonnegative real number you get if you remove the j. we enclose it between vertical lines. the value of the imaginary number increases. just as we do with real numbers and real-number expressions. and also remove the minus sign (if there is one). If you go downward. which is identical to the real number zero. the value decreases. For example.352 Imaginary and Complex Numbers
Larger positively Smaller positively
j3 j2 j j0 = 0 -j -j 2 Smaller according to the traditional definition “Number reflector” plane Larger according to the traditional definition
Smaller negatively Larger negatively
-j3
Figure 21-1 The imaginary number line. as well as the definitions for “larger than” and “smaller than. To denote the absolute value of an imaginary number or imaginary-number expression.

as shown on the left side of Fig. 21-2. move upward Think of upward distances on the imaginary number line as positive imaginary displacements. jb1 + jb2. We start at the point for −j 3 and move up 2 units. then |jb| = b and |−jb| = b
To add. We’re talking about displacement here. As an example. we move downward by the equivalent distance. On the right. a number-line rendition of
−j 3 + j 2. That gets us to the point for −j 3 + j 2. not simple distance.The Imaginary Number Line
353
In general. so negatives can make sense! An upward displacement
j3 j2 j 0 Finish here Move upward by 2 units Start here -j -j 2 -j3 Finish here Start here
Move upward by -3 units
Figure 21-2 On the left. a number-line rendition of j 2 + (−j 3). and downward distances as negative imaginary displacements. It happens to be −j. That will get us to the point representing the sum of the two numbers. Then we move up along the line by b2 units. When we move negatively upward. if b is any nonnegative real number. we first find the point on the number line representing jb1.
. Now suppose that we start with j2 and travel upward by −3 units. suppose b1 = −3 and b2 = 2. If we have an imaginary number jb1 and we want to add another imaginary number jb2 to it.

familiar with respect to multiplication and addition of real numbers. we end up at the same point. We have geometrically analyzed these two facts: −j 3 + j 2 = −j and j 2 + (−j 3) = −j
To subtract. we move negatively downward along the imaginary number line. We add a negative imaginary number to some other imaginary number.354 Imaginary and Complex Numbers
of −3 units is the same as a downward displacement of 3 units. meaning that we actually travel upward.
Solution
When we subtract a negative imaginary number. Then we travel downward by b2 units along the imaginary number line. When we add the imaginary numbers −j3 and j2 in either order as shown. Write it down in the simplest possible form. Adding a negative imaginary number is the same thing as subtracting the product of −1 and that number. Let’s see how the distributive law. move downward Look again at the right-hand side of Fig. and not just to j itself? If so. which corresponds to −j. That will get us to the point representing jb1 − jb2. We start at −j3 and
. 21-2. This process is shown on the right in Fig. 21-2. from j and then find the sums like this: −j 3 + j 2 = j (−3 + 2) = j (−1) = −j and j 2 + (−j 3) = j [2 + (−3)] = j (2 − 3) = j (−1) = −j
Here’s a challenge!
In terms of the imaginary number line. express the fact that when we subtract −j 5 from −j 3.
Are you confused?
Do you suspect that the laws of real-number arithmetic apply to all imaginary numbers. we get j 2. we must first find the point on the number line representing jb1. can be used to scrutinize the two imaginary-number sums we just analyzed. We can separate the real-number multiples. Figure 21-3 shows how this works. 9 if you want to review those laws. called the real coefficients. If we have an imaginary number jb1 and we want to subtract another imaginary number jb2 from it. you’re right! Look back at the end of Chap.

we put down the real-number part first. the term “complex” does not mean “complicated.Real + Imaginary = Complex
355
Figure 21-3 Here. In this context.” A better word would be “composite. we go upward by the equivalent distance.
How they are written When we write a complex number. then a plus or minus sign.” but that term has already been taken! (A composite number is a natural number that can be factored into a product of two or more primes. 9 apply to complex numbers. which means we really move up by 5 units. When we go negatively downward. We finish at the point corresponding to j 2. ending up with j2. Here are some examples:
4 + j3 −4 + j 5
. We can write −j 3 − (−j 5) = j 2 Simplifying. you get a complex number. and expressions containing complex numbers.
j3 j2 j 0 -j -j 2 -j3 Start here Move downward by -5 units Finish here
move down by −5 units. we start
with −j3 and then subtract −j5. complex-number variables.) All the rules of arithmetic you learned in Chap. and the imaginary-number part. we can write −j 3 + j 5 = j 2
Real + Imaginary = Complex
When you add a real number and an imaginary number.

They are also identical to the imaginary numbers j 0 and −j 0. showing
five values plotted as points. The third and fourth of the above complex numbers can be converted to the sums −5 + (−j3) and 1 + (−j 6) The complex numbers 0 + j 0 and 0 − j 0 are the same as the real number 0.-j6 ) 2 4 6 Origin = 0 + j 0 =0 (4. The dashed reference lines help to show the coordinates of the points on the axes. The set of coordinates shown in Fig. we can rewrite it as a sum.356 Imaginary and Complex Numbers
−5 − j 3 1 − j6 0 + j0 When a complex number is written as a difference between a real number and an imaginary number.+j3)
Figure 21-4 The complex-number plane. in which we can plot
jb j6 (-4. 21-4 is the complex-number plane.
.-j3) -j 4 -j 6 (1.+j5) j4 j2 a -6 -4 -2 -j 2 (-5.
The complex-number plane The set of complex numbers needs two dimensions—a plane—to be graphically defined.

Any point in the plane. The real part is expressed toward the right for positive and toward the left for negative. You’ll work with them a lot in the coming chapters. the point is in the third quadrant. For example. You won’t have to do this very often.” Any sum with two addends is a binomial. but if you ever find yourself faced with
. a complex number a + jb is called pure imaginary. If a is positive and b is negative. it’s best to convert a difference to a sum. the sum of 4 + j 7 and 45 − j 83 works out like this:
(4 + j 7) + (45 − j 83) = (4 + 45) + j(7 − 83) = 49 + j(−76) = 49 − j 76 Subtracting complex numbers is a little more involved. like this: (4 + j 7) − (45 − j83) = (4 + j 7) + [−1(45 − j83)] = (4 + j 7) + (−45 + j83) = −41 + j 90
Multiplying complex numbers When you multiply one complex number by another. and d are real numbers. can be expressed as an ordered pair (a. you should treat both of the numbers as sums called binomials. b. c. which means “expression with two names. For example. representing a unique complex number. If a ≠ 0 and b = 0. If both a and b are negative. If a is negative and b is positive. then
(a + jb)(c + jd ) = ac + jad + jbc + j 2bd = (ac − bd ) + j(ad + bc) Remember that j 2 = −1! That’s why you get a minus sign between ac and bd. we can find (4 + j7) − (45 − j 83) by multiplying the second complex number by −1 and then adding the two complex numbers. the point is in the second quadrant. the point representing a complex number is in the first quadrant of the plane. things get a little messy. we must add the real parts and the complex parts separately.jb) or written as a + jb. The imaginary part goes upward for positive and downward for negative. where a and b are real numbers and j is the unit imaginary number. If both a and b are positive.Real + Imaginary = Complex
357
quantities that are part real and part imaginary. If a = 0 and b ≠ 0. a complex number a + jb is called pure real. If a.
Dividing complex numbers When you want to divide a complex number by another complex number. 9.
Adding and subtracting complex numbers When we want to add two complex numbers. This rule is an adaptation of the product of sums rule you learned in Chap. the point is in the fourth quadrant.

the absolute value is −(−34) = 34. Therefore. We can write this as a formula:
|a + jb| = (a2 + b2)1/2
.
Absolute value of a complex number The absolute value of a complex number a + jb is the distance from the origin (0. and d are real numbers. like this:
a + jb and a − jb These two numbers are called complex conjugates. b. the absolute value of 0 + jb is equal to −b. Finally. Then we add the squares. jb) on the complex-number plane. we take the positive square root of that sum. It is the same as −22 + j 0. This is a pure real number. For a + j 0 when a is positive. and if c and d are not both equal to 0. For a pure imaginary number 0 + jb where b is positive. the absolute value is b. you can use the following formula. because j0 = 0.358 Imaginary and Complex Numbers
a complex-number division problem. then (a + jb)/(c + jd ) = (ac + bd )/(c 2 + d 2) + j(bc − ad )/(c 2 + d 2)
Complex conjugates Every complex number has a sort of “mirror image. If a. we square both a and b separately. j 0) to the point (a. You’ll see some examples in Chap. the absolute value of this complex number is −(−22) = 22. The value of b is −34. because 0 − j34 = 0 + j(−34). but where one is a sum and the other is a difference. For a + j 0 when a is negative. When you add a + jb to its conjugate. you get a pure real number: (a + jb) + (a − jb) = (a + a) + (jb − jb) = 2a + j 0 = 2a When you multiply a + jb by its conjugate. If b is negative. First. the absolute value is −a.” Imagine two complex numbers that have the same real coefficients. Therefore. What about the absolute value of 0 − j34? This is a pure imaginary number. the absolute value must be found by going through a little arithmetic. c. If a complex number a + jb is neither pure real or pure imaginary. Suppose we want to find the absolute value of −22 − j 0. the absolute value is a. 23. you get another pure real number: (a + jb)(a − jb) = a2 − jab + jba − j 2b2 = a2 − jab + jab + b2 = a2 + b2 Complex conjugates show up together when you solve certain equations.

rational numbers. sometimes denoted C. you might take courses that take you “farther out. 21-5 is a parallelogram. j 3) can be found by taking the ratio of the difference in jb to the difference in a. j 0) and (2.Real + Imaginary = Complex
359
Let’s find the absolute value of 3 − j 4. These three points. The four points lie at the vertices of a parallelogram. as well as their sum. To find the sum of these two complex numbers. along with the origin. The set of complex numbers. It’s a special sort of quadrilateral. Someday. and real numbers? The answer is that every one of those other sets is a proper subset of the set of complex numbers. form the vertices of a quadrilateral (four-sided geometric figure). like this: (2 + j 3) + (3 + j ) = (2 + 3) + j (3 + 1) = 5 + j4 Figure 21-5 shows the two complex numbers. Squaring both of these and adding the results gives us 32 + (– 4)2 = 9 + 16 = 25 The positive square root of 25 is 5. along with their sum and the origin. In this case.
Here’s a challenge!
Find the sum (2 + j3) and (3 + j ). imaginary numbers. integers. It’s as far into the universe of numbers as we’ll go. like this: m1 = Δjb /Δa = ( j 3 − j 0)/(2 − 0) = j 3/2
. in the complexnumber plane. irrational numbers. Then plot both of these numbers. |3 − j4| = 5. That’s why the j is all alone in the second addend above.
Are you confused?
Do you wonder how the set of complex numbers relates to the sets of natural numbers. To prove that the quadrilateral in Fig. we can show that the pairs of line segments forming opposite sides have identical slopes. What sort? Why?
Solution
First. The slope m1 of the line segment connecting (0. we add their pure real and pure imaginary parts separately and then put the results back together. there’s no need to write down the numeral 1 after the j. Therefore. a = 3 and b = −4.” Who knows? Maybe you’ll discover or invent a new realm of numbers that nobody has worked with before. we should note that when j is multiplied by 1 in a complex number. Remember from geometry: A parallelogram is a quadrilateral in which the pairs of opposite sides are parallel. as ordered pairs in the complex-number plane. is the “grandmother of all number sets” as far as most algebra is concerned.

by all means try it! 1. Second.
Practice Exercises
This is an open-book quiz. 4. If you think you can solve a particular problem in a quicker or better way than you see there. subtract the second from the first. First. Here’s a hint: Note that j −1 = j 3−4.Practice Exercises
361
Those two opposite sides are also parallel. j −5. use the difference of powers law.
. Then subtract the first from the second. The laws of arithmetic for real numbers also apply to imaginary numbers. 6. This proves that the quadrilateral in Fig. and d are real numbers. The solutions in the appendix may not represent the only way a problem can be figured out. prove this. First. c. b. Using what we’ve learned in the chapter text and so far in this set of exercises. Find the following: (a) (4 + j 5) + (3 − j 8) (b) (4 + j 5) − (3 − j 8) (c) (4 + j 5)(3 − j 8) (d) (4 + j 5)/(3 − j 8) 7. Determine the value of j −1 in two ways. Using the difference of powers law and all the other things we’ve learned. C. or else all four points will lie along a single straight line (a “squashed parallelogram”). Find the difference between the complex conjugates (a + jb) and (a − jb). You may (and should) refer to the text as you solve these problems. here’s a hint: Find the value of an unknown (call it z) when 1/j = z /1. If you’re ambitious. What is the value of j −2? The value of j −4? The value of j −6? The value of j −8? What happens as this trend continues? 3. determine the values of j −3. where a. You’re on your own. Here’s a hint: Call the two complex numbers a + jb and c + jd. use the law of cross multiplication. 21-5. Here are some hints: j −3 = j 1−4 j −5 = j −1−4 j −7 = j −3−4 What happens as this trend continues? 5. you’ll get a parallelogram.
Here’s an extra-credit challenge!
Whenever you add two complex numbers and diagram the process after the fashion of Fig. 21-5 is a parallelogram. On that basis. how can we determine the value of j 0? 2. create a table that shows what happens when j is raised to any integer power. Again. Don’t hurry! You’ll find worked-out answers in App. and j −7.

Suppose k is a positive real-number constant. 9. Find the quotient (a − jb)/(a + jb). assuming that a and b are not both equal to 0.
. assuming that a and b are not both equal to 0. Find the quotient (a + jb)/(a − jb). How many pure real numbers have absolute values equal to k ? How many pure imaginary numbers have absolute values equal to k? How many complex numbers have absolute values equal to k ? Draw a diagram in the complex-number plane that shows these situations. How does this compare with the answer to Exercise 8? 10.362 Imaginary and Complex Numbers
8.

it isn’t in polynomial standard form. First. But the first constant. All of the following are quadratic equations in polynomial standard form: 3x 2 + 2x + 5 = 0 3x 2 − 4x = 0 −7x 2 − 5 = 0 −4x 2 = 0 In all but the first of these. because the term containing x 2 vanishes: bx + c = 0
Mutant quadratics Quadratic equations frequently appear in disguise. the one by which x 2 is multiplied. Here’s the general equation:
ax 2 + bx + c = 0 where a. you find 0 all by itself. Second. the left side of the equals sign contains a constant multiple of the variable squared. If you set a = 0 in the standard form of a single-variable quadratic equation. it can be morphed into that form without changing the set of roots we get when we solve it. you’ll get the standard form for single-variable first-degree equation. some of the constants equal 0. no matter how complicated it looks when you first see it. The above expressions can be written as pure sums like this: x + (−3) −x + 3x + (−6) x + (−y 3) + 7z 2 2x + (−2y 5) + (−2z 7) ax 4y + (−bxz 3) + (−cy 2z 2)
2
Polynomial standard form Any single-variable quadratic. Here are some examples: x 2 = 2x + 3 x = 4x 2 − 7 x 2 + 4x = 7 + x x − 2 = −8x 2 − 22 3 + x = 2x 2
. can never be 0 in a quadratic equation. and c are constants. added to a constant. b. and x is the variable. all cases of subtraction are sums in disguise.364 Quadratic Equations with Real Roots
As you know. can be rewritten so it appears in polynomial standard form. When the equation is in this form. On the right side of the equals sign. I call them mutant quadratics. added to a constant multiple of the variable itself. Every such equation has two things in common.

These terms are: −23x 2/(−10) and 6x 2/5 We can add these two terms to get a single term in x 2. and finally subtract 4 from each side. or it isn’t a quadratic. but you aren’t sure. and if we multiply both the numerator and denominator of the second expression by 2. If you subtract 4x 2 from each side. As things work out. You think it’s a quadratic. but nothing beyond the mathematical skill we’ve acquired by now! Let’s look closely at the right side of this equation. That can be tricky. If you can’t convert it.366 Quadratic Equations with Real Roots
Are you confused?
Suppose you see an equation in one variable. then two things are possible: you didn’t try hard enough. you can be certain that it’s a quadratic. you must morph the equation into a form that’s obviously not convertible into polynomial standard form. not a quadratic: 7x 2/2 + 7x − 5 = −23x 2/(−10) + 6x 2/5 + 3x
Solution
It takes a little intuition to solve this. then subtract 7x from each side. you get −x 2 − 6x − 12 = 0
Here’s a challenge!
Show that the following is the equivalent of a first-degree equation. If it isn’t a quadratic and you want to prove that it isn’t. we get 23x 2/10 and 12x 2/10 We now have a common denominator. Here’s an example: 3x 2 + x − 8 = 4x 2 + 7x + 4 If you can convert this equation to polynomial standard form. It contains two terms with x 2. The next “challenge” will show you an example. you can convert the above equation to polynomial standard form. If we multiply both the numerator and denominator of the first of the above expressions by −1. so we can find the sum 23x 2/10 + 12x 2/10 = 35x 2/10
.

To produce a quadratic. This allows us to rewrite the right side of the original equation to obtain 7x 2/2 + 7x − 5 = 7x 2/2 + 3x Subtracting 7x 2/2 from each side.Binomial Factor Form
which reduces to 7x 2/2.
Binomials in quadratics When a left side of a quadratic is expressed as a product of binomials. The multiplicand in that term is the coefficient of the variable. As you’ll soon see. As long as p. there’s no restriction on the values of the coefficients and constants. Other than that. The second term is a constant. we get a quadratic equation. this form is simpler than the polynomial standard form. first-degree form: 4x − 5 = 0
Binomial Factor Form
There’s another way to express a quadratic equation: as a product of two binomials that is equal to 0. we get 7x − 5 = 3x
367
When we subtract 3x from each side now. sometimes called the stand-alone constant.
. The first term in each binomial is a multiple of the variable. the binomial factor form of a quadratic tells you the roots directly. we get an equation in the standard single-variable. r. neither of the coefficients p nor r can be equal to 0. In some ways. Here’s the general binomial factor form for a quadratic:
(px + q)(rx + s) = 0 where p and r are the coefficients. and s are all real numbers. and if we then set the product equal to 0. Here are some examples:
x+1 x−5 3x + 5 −17x + 24 8x − 13 −7x − 11
Multiplying two binomials If we multiply two binomials of the above sort where both binomials contain the same variable. then the resulting equation will have real roots. both of the binomials must be in a specific form. and x is the variable. q and s are the constants. q.

and you have the one-variable first-degree equation (ps + qr)x + qs = 0
What are the roots? If we find a quadratic equation in binomial factor form and we want to find the roots. Start with the expression (px + q)(rx + s) Multiplying this out according to the product of sums rule gives us pxrx + pxs + qrx + qs Using the commutative law for multiplication in the first two terms and then simplifying the first term.368 Quadratic Equations with Real Roots
The fact that the above general equation is a quadratic might not appear obvious. for a quadratic in the variable x :
(px + q)(rx + s) = 0
. the reason for this restriction should be clear now. but when we multiply out the left side of the equation using the product of sums rule. we’re in luck! Here is the binomial factor form again. we get prx 2 + psx + qrx + qs The distributive law “in reverse” allows us to rewrite this as prx 2 + (ps + qr)x + qs When we set this equal to 0. we get an equation in polynomial standard form: prx 2 + (ps + qr)x + qs = 0
Are you confused?
If you have trouble seeing why the above equation is in polynomial standard form. you get ax 2 + bx + c = 0 If you wonder about the requirement that p ≠ 0 and r ≠ 0 in the binomial factor form. you can rename the coefficients and constants like this: pr = a ps + qr = b qs = c By substitution. then pr = 0. If p = 0 or r = 0. the term containing x 2 vanishes. the truth is revealed.

. The equation then reduces to 0 = 0. If X is the set of solutions. The solution set for a true quadratic. we can subtract s from each side to obtain rx = −s Then we divide through by r. getting x = −s /r The roots of the quadratic are x = −q /p or x = −s /r. If either factor (that is. As you continue studying algebra. you’ll come across equations with solution sets containing three.
Here’s a challenge!
Consider the following quadratic equation in standard form: x 2 − 2x − 15 = 0 Put this equation into binomial factor form. which we have said is nonzero. or more elements. then X = {−q /p.Binomial Factor Form
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where neither p nor r equals 0. we can subtract q from each side. indicating a root! The roots of the above quadratic can therefore be found by solving these two first-degree equations: px + q = 0 and rx + s = 0 In the first equation. called the solution set.−s /r} The solution set for an equation with multiple roots is the set containing all of those roots. which we have restricted to nonzero values. then the entire expression on the left side of the equals sign becomes 0. Then find the solution set X. either binomial) happens to equal 0. never has more than two elements. Here’s a hint: The coefficients and constants in the binomial factor form are integers. getting px = −q Dividing through by p. we get x = −q /p In the second equation. four. however.

not into binomials with real-number constants and coefficients. That leaves us with these choices: • • • • q = 15 and s = −1 q = −15 and s = 1 q = 3 and s = −5 q = −3 and s = 5
Now let’s look at the coefficient of x in the original equation. Remember the general binomial factor form: (px + q)(rx + s) = 0 We must find the values p. others are difficult. we have eight possibilities: • • • • • • • • q = 15 and s = −1 q = −15 and s = 1 q = 1 and s = −15 q = −1 and s = 15 q = 3 and s = −5 q = −3 and s = 5 q = 5 and s = −3 q = −5 and s = 3
We can trim this down to four possibilities by eliminating duplicate scenarios. q = 3. we have two possibilities: • p = 1 and r = 1 • p = −1 and r = −1 Because qs = −15. We get (px + q)(rx + s) = (x + 3)(x − 5) = x 2 − 5x + 3x − 15 = x 2 − 2x − 15
. and s. we get ps + qr = 1 × (−5) + 3 × 1 = −5 + 3 = −2 That’s the right coefficient for x in the original! Let’s try those numbers in the binomial factors for the left side of the equation. q. Because pr = 1. This tells us that qs = −15. we’ll see that if we let p = 1. This tells us that ps + qr = −2 If we “play around” with our choices for awhile. Learning how to factor a quadratic takes some practice. not surprisingly. and some can’t be factored at all—at least. That means pr = 1 in the binomial factor form. r = 1. factoring. We’re told that they’re all integers. The last constant in the original equation is −15. The coefficient of the x 2 term in the original equation is equal to 1.370 Quadratic Equations with Real Roots
Solution
The process of putting a quadratic into binomial factor form is called. Some quadratics are easy to factor. r. It’s −2. and s = −5.

we have x 2 − 2x − 15 = 0 (−3)2 − 2 × (−3) − 15 = 0 9 − (−6) − 15 = 0 9 + 6 − 15 = 0 15 − 15 = 0 0=0 For x = 5. One of these methods is called completing the square.5} Let’s check to be sure these solutions work in the original equation. the left side of the equation becomes equal to 0. For x = −3. A polynomial of this type is called a perfect square. If x = −3 or x = 5. Here are some examples:
• • • • The polynomial x 2 + 2x + 1 factors into (x + 1)2 The polynomial x 2 − 2x + 1 factors into (x − 1)2 The polynomial x 2 + 4x + 4 factors into (x + 2)2 The polynomial 9x 2 + 12x + 4 factors into (3x + 2)2
Quadratics built from perfect squares (by setting the right side of the equation equal to 0) are easy to solve. The solution set X is therefore X = {−3. so in effect there’s only one root. Now we know that it can be written as (x + 3)(x − 5) = 0
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The roots are easy to find. we have x 2 − 2x − 15 = 0 52 − 2 × 5 − 15 = 0 25 − 10 − 15 = 0 15 − 15 = 0 0=0
Completing the Square
There are other ways to look for the roots of a quadratic.”
.Completing the Square
That’s the left side of the original quadratic. “done twice over. equivalent to a single factor multiplied by itself. The two roots are identical.
Perfect squares Suppose that we come across a quadratic equation whose left side breaks down into two identical factors.

This gives us the following first-degree equations. respectively: x+1=0 x−1=0 x+2=0 3x + 2 = 0 The solutions to these first-degree equations are easy to find: x = −1 x=1 x = −2 x = −2/3 These are the roots of the original quadratics. {1}.372 Quadratic Equations with Real Roots
Such a root is said to have multiplicity 2. {−2}.
Positive numbers on the right Now we’ll depart slightly from the polynomial standard form. Let’s use the same perfect squares as above. If the above polynomial expressions are placed on the left sides of equations with 0 on the right. We can write down the solution sets as the single-element sets {−1}. and {−2/3}. Here are the new equations:
x 2 + 2x + 1 = 1 x 2 − 2x + 1 = 4 x 2 + 4x + 4 = 16 9x 2 + 12x + 4 = 25
. we get these quadratics: x 2 + 2x + 1 = 0 x 2 − 2x + 1 = 0 x 2 + 4x + 4 = 0 9x 2 + 12x + 4 = 0 Respectively. but set the right sides to positive numbers rather than 0. we can take the square roots of both sides in each case without “plus-or-minus” ambiguity. they factor into: (x + 1)2 = 0 (x − 1)2 = 0 (x + 2)2 = 0 (3x + 2)2 = 0 Because the expressions on the left sides of the above equations are equal to 0.

(Recognizing perfect squares when they appear in polynomial form is a “sixth sense” that evolves over time. but as long as the left side is a perfect square. we obtain (x + 1)2 = 1
. getting
x 2 + 2x = 0 We can take the second equation and subtract 4 from each side. getting x 2 + 2x + 1 = 1 That gives us a perfect square on the left side. getting x 2 − 2x − 3 = 0 In the third equation. and it takes practice to develop it. In this way. we can solve the equation just as we did in the four cases above. We can take the first quadratic from the above list and subtract 1 from each side. getting a perfect square on the left side.
Are you confused?
Now that we’ve taken four solutions and manufactured four problems from them. we can get a good “feel” for how completing the square actually works. we can sometimes add a positive real number to both sides. let’s retrace our steps and get the solutions back. you’ll get quadratics with perfect squares on their left-hand sides.374 Quadratic Equations with Real Roots
x 2 + 4x + 4 = 16 9x 2 + 12x + 4 = 25
Morphing into a perfect square When we come across a quadratic equation in polynomial standard form. in the fourth equation.) Factoring. That will leave us with a nonzero value on the right. Imagine that we’re confronted with the following four quadratics in polynomial standard form: x 2 + 2x = 0 x − 2x −3 = 0
2
x 2 + 4x − 12 = 0 9x 2 + 12x − 21 = 0 We can take the first of these equations and add 1 to each side. we can subtract 16 from each side to obtain x 2 + 4x − 12 = 0 Finally. we can take 25 away from each side and get 9x 2 + 12x − 21 = 0 If you add the right constants to both sides of each of these equations.

The Quadratic Formula
The technique of completing the square can be applied to the general polynomial standard form of a quadratic equation. and c are realnumber constants with a ≠ 0. getting a negative number on the right side? That’s a good question. so the solution set is {0.
Are you still confused?
Do you wonder what happens if.−2}.
Deriving the formula Remember the polynomial standard form where x is the variable. This gives us a tool for solving quadratics by “brute force”: the so-called quadratic formula. The other three equations can be worked out in similar fashion. We’ll deal with such equations in Chap. The general formula is
ax 2 + bx + c = 0 Let’s rewrite this as ax 2 + bx = −c
. you must subtract a positive number from both sides. and fourth quadratics from above: x 2 − 2x −3 = 0 x 2 + 4x − 12 = 0 9x 2 + 12x − 21 = 0
Solution
You’re on your own! Start with perfect squares on the left sides of the equals signs and positive numbers on the right.The Quadratic Formula
We can take the square root of both sides and get x + 1 = ±1 which can be expressed as the pair x + 1 = 1 or x + 1 = −1
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The solutions are found to be x = 0 or x = −2. but with the second. and a.
Here’s a challenge!
Go through maneuvers similar to those we just completed. in order to complete the square in a quadratic. the roots turn out to be imaginary or complex. 23. In that case. third. b. and then take away those positive numbers from both sides to “unsquare” the equations.

we can divide each side by a. getting x 2 + (b /a)x = −c /a It’s tempting to think that there must be some constant that we can add to both sides of this equation to get a perfect square on the left side of the equals sign. we can simplify the expression on that side of the equals sign a little bit. remembering the negative as well as the positive. we get x + b /(2a) = ±[(b 2 − 4ac) / (4a 2)]1/2 The denominator in the right side is a perfect square. considering it as a ratio of square roots rather than the square root of a ratio. but that constant does exist. we obtain x 2 + (b /a)x + b 2/(4a 2) = −c /a + b 2/(4a 2) We can now factor the left side into the square of a binomial to get [x + b /(2a)]2 = −c /a + b 2/(4a 2) The two terms in the right side of this equation can be added using the sum of quotients rule from Chap.
. and c (finally!). so the equation becomes [x + b /(2a)]2 = (b 2 − 4ac) / (4a 2) If we take the square root of both sides here. we get [x + b /(2a)]2 = (−4ac + b 2) / (4a 2) Let’s rewrite the numerator on the right side as a difference. we get x = ±(b 2 − 4ac)1/2 / (2a) − b /(2a) which expresses x in terms of the constants a. It takes some searching. An equation that states the general solution to an unknown is called a formula. it’s equal to (2a)2. 9 to obtain [x + b /(2a)]2 = (−4a 2c + ab 2) / (4a 3) Canceling out the extra factors of a in the numerator and denominator on the right side. When we add it to both sides of the above equation. We obtain x + b /(2a) = ±(b 2 − 4ac)1/2 / (2a) If we subtract b /(2a) from both sides. b.376 Quadratic Equations with Real Roots
Because a ≠ 0 in any quadratic equation. Therefore. It is b 2/(4a 2).

We’ll explore situations like this in the next chapter. That’s an imaginary number! A negative discriminant gives us imaginary or complex roots. b 2 − 4ac = 0.
Are you confused?
Do you wonder how we came up with the constant b 2/(4a 2) to make a perfect square in the process of deriving the quadratic formula? Again. we made a “quantum leap” when we claimed that the polynomial x 2 + (b /a)x + b 2/(4a 2) is a perfect square that can be factored into [x + b /(2a)]2 Show that this is actually true. this is the “sixth sense” at work. If the discriminant is equal to 0. This gives us a quick way to tell whether a quadratic equation has two roots or only one. then the associated quadratic has two real roots. In that case.
. There are plenty of quadratic equations in which the discriminant is a negative real number. We’ll also see what happens when one or more of the coefficients in a quadratic equation are imaginary or complex. The discriminant here is b 2 − 4ac = 42 − 4 × 4 × 36 = 16 − 576 = −560 When we apply the quadratic formula. The only difference is that the second coefficient is 4 rather than 24. Here’s an example: 4x 2 − 4x + 36 = 0 This is almost exactly the same equation as we solved in the second example above.
Here’s a challenge!
In the derivation of the quadratic formula. we must take the square root of −560. The quantity b 2 − 4ac is called the discriminant of the general quadratic equation
ax 2 + bx + c = 0 If the discriminant is a positive real number. a form of intuition that you can develop only with practice. then the quadratic has one real root with multiplicity 2. it doesn’t matter whether we add (b 2 − 4ac)1/2 to −b or subtract (b 2 − 4ac)1/2 from –b in the formula.378 Quadratic Equations with Real Roots
The discriminant In the second example above.

Then find the roots and state the solution set. 2x 2 + 8x − 10 = 0 4. C.Practice Exercises
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Solution
We can work it the other way. by all means try it! 1. Don’t hurry! You’ll find worked-out answers in App. x 2 + 10x + 25 = 0 3. Factor the following quadratic. Here’s a hint: The coefficients and constants in the factors are all integers. getting x 2 + 2xb /(2a) + b 2/(4a 2 ) Canceling out the 2 in the numerator and denominator of the middle term. The solutions in the appendix may not represent the only way a problem can be figured out. You may (and should) refer to the text as you solve these problems. If you think you can solve a particular problem in a quicker or better way than you see there. Factor the following quadratic. we obtain x 2 + xb /a + b 2/(4a 2) which can also be written as
x 2 + (b /a)x + b 2/(4a 2 )
That’s the original polynomial. this becomes x 2 + xb /(2a) + xb /(2a) + b 2/(4a 2 ) We can add the two middle terms together. Factor the following quadratic.
Practice Exercises
This is an open-book quiz. 12x 2 + 7x − 10 = 0
. Here’s a hint: The coefficients and constants in the factors are all integers. like this: [x + b /(2a)] [x + b /(2a)] When we apply the product of sums and product of quotients rules from Chap. Here’s a hint: The coefficients and constants in the factors are all integers. Multiply out the following equation. putting it into polynomial standard form. multiplying the factors out to get the polynomial. Then find the roots and state the solution set. Then find the roots and state the solution set. 9. (−7x − 5)(−2x + 9) = 0 2. Let’s rewrite the above squared binomial as a product of sums.

Manipulate the following equation so it has the square of a binomial on the left side of the equals sign and 0 on the right side. Here’s a hint: The coefficient and constant in the binomial are both integers. Manipulate the following equation so it has the square of a binomial on the left side of the equals sign and a positive real number on the right side. using the quadratic formula.380 Quadratic Equations with Real Roots
5. How many real roots does the following quadratic have? Find the real root or roots. What is the real-number solution set? −2x 2 + 3x + 35 = 0 10. What is the single root of the quadratic stated in Prob. 16x 2 − 40x + 25 = 0 6. Here’s a hint: The coefficient and constant in the binomial are both integers. 5? What is the solution set? 7. What is the real-number solution set? 4x 2 + x + 3 = 0
. if any exist. How many real roots does the following quadratic have? Find the real root or roots. What are the roots of the quadratic stated in Prob. using the quadratic formula. x2 + 6x − 7 = 0 8. 7? What is the solution set? 9. if any exist.

But we must never use it when d = 0 or when d > 0. and the negative square root is −j (471/2).382 Quadratic Equations with Complex Roots
Let’s take an example. we have ±(d 1/2) = ± j (|d |1/2) Substituting ±j (|d |1/2) in place of ±(b 2 − 4ac)1/2 in the quadratic formula. is negative. and we find that d = −16. we get x = [−b ± j (|d |1/2)] / (2a) This equation can be used if and only if the real-number discriminant. so d = b 2 − 4ac = 12 − 4 × 4 × 3 = 1 − 48 = −47 The positive square root of −47 is j (471/2). b = 1. and the negative square root of d is equal to −j 4. the j operator does not belong there. a = 4. then |d | > 0. then |d | = −d. Then the positive square root of d is equal to j 4.) We can express the positive square root of d as d 1/2 = j (|d |1/2) and we can express the negative square root of d as −(d 1/2) = −j (|d |1/2) Stated as a “plus-or-minus” expression.
Imaginary roots: a specific case Consider the following quadratic equation in which the coefficient of x 2 is positive. 22: 4x 2 + x + 3 = 0 In this case. Suppose that we work out the discriminant d for a quadratic. the coefficient of x is equal to 0. and c = 3. Now let’s look at the general case. If d < 0. d. it also happens to be true that if d < 0. Here’s another example. because in those cases. and the stand-alone constant is positive:
3x 2 + 75 = 0
. It’s important to remember what “if and only if ” means in this context! We can always use this formula when d < 0. (As you know. a fact that we’ll use later in this chapter. Let’s revisit the quadratic stated in Practice Exercise 10 at the end of Chap.

we get 3x 2 = −75 Dividing through by 3 gives us x 2 = −25 When we take the positive-or-negative square root of both sides. b = 0. and c = 75. The discriminant is therefore d = b 2 − 4ac = 02 − (4 × 3 × 75) = −900 The positive-or-negative square root of the discriminant is ±(d 1/2) = ±j (|−900|1/2) = ±j (9001/2) = ±j 30 The roots can now be found as x = [−b ± j (|d |1/2)] / (2a) = (0 ± j 30) / (2 × 3) = ±j 30/6 = ±j 5
Are you astute?
Do you suspect that this particular equation can be solved more easily without resorting to the quadratic formula? If so. you’re right! If we subtract 75 from each side. If we write it in polynomial standard form. we get
ax 2 + 0x + c = 0
. we obtain x = ±[(−25)1/2] = ±j 5
Imaginary roots: the general case Let’s see what happens with a general quadratic when the coefficient of x is 0.Complex Roots by Formula
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In the general polynomial standard form. we have a = 3.

Therefore.384 Quadratic Equations with Complex Roots
which simplifies to ax 2 + c = 0 Subtracting c from each side gives us ax 2 = −c We can divide through by a because. Here’s an example:
(45/2)x 2 + 3x + 1 = 0 In this case. in a quadratic. and c = 1. the coefficient of x. is not equal to 0. then −c /a is negative. a is never 0. Therefore. Doing that. b = 3. then −c / a is positive. That means the roots are pure imaginary and are additive inverses of each other: x = j (|−c /a|1/2) or x = −j (|−c /a|1/2) In this case.
Conjugate roots The discriminant in a quadratic can be negative even when b. the roots are both real and are additive inverses of each other: x = (−c /a)1/2 or x = −[(−c /a)1/2] If a and c have the same sign. d < 0. so we have 4ac = 4 × (45/2) × 1 = 90
. we obtain x = ±[(−c /a)1/2] If a and c have opposite signs. we get x 2 = −c /a Taking the positive-or-negative square root of each side. a = 45/2. −4ac < 0. The only requirement is that 4ac be larger than b 2. the discriminant is d = b 2 − 4ac = 02 − 4ac = −4ac Because a and c have the same sign.

then the roots are not pure imaginary.
. The solution set X in this example is X = {−1/15 + j (1/5). but if you’re careful and patient. all the “garbage” will drop out in the end. In any quadratic: • If the discriminant is a negative real number and the coefficient of x is 0. d = b 2 − 4ac = 9 − 90 = −81 This tells us that the roots of the quadratic are not real numbers. Here’s a hint: This is a messy process. • If the discriminant is a negative real number and the coefficient of x is a nonzero real number. we can plug in the value −81 for d in the “abbreviated discriminant” form of the quadratic formula: x = [−b ± j (|d |1/2)] / (2a) = [−3 ± j (|−81|)1/2] / [2 × (45/2)] = [−3 ± j (811/2)] / 45 = (−3 ± j 9) / 45 = −3/45 ± j (9/45) = −1/15 ± j (1/5) The roots are x = −1/15 + j (1/5) or x = −1/15 − j (1/5) These are complex conjugates. As things work out. the roots are always complex conjugates in a quadratic where d < 0. and are additive inverses. then the roots are pure imaginary.
Are you confused?
The above derivations are abstract. Here are the results. plug the roots we’ve just found into the original quadratic to be sure that they work.Complex Roots by Formula
385
and b 2 = 32 =9 Therefore. To find the roots. −1/15 − j (1/5)}
Are you ambitious?
For complementary credit. but are complex conjugates. even when b ≠ 0. wrapped up into two statements. but it’s important that you follow through them so you understand the reasoning behind each step. You’re on your own.

“Are p and q really both nonzero?” The answer is “Yes.” We can be sure that p ≠ 0 because b ≠ 0 and a ≠ 0. so −b /(2a) can’t be 0.
Solution
Consider once again the general polynomial equation ax 2 + bx + c = 0 The discriminant is d = b 2 − 4ac Suppose that d < 0 and b ≠ 0. To bring things into focus. like this: x = −b /(2a) ± j (|d |1/2)/(2a) Therefore.386 Quadratic Equations with Complex Roots
Here’s a challenge!
Prove the second of the above statements. Let p = −b /(2a) and q = (|d |1/2)/(2a) Now we can rewrite the roots as x = p + jq or x = p − jq These are complex conjugates (but not pure imaginary numbers) for all nonzero real numbers p and q. we can split that expression into a sum or difference of two ratios with a common denominator. we can change a couple of names.
. so (|d |1/2)/(2a) can’t be 0. we can write the two roots as x = −b /(2a) + j (|d |1/2)/(2a) or x = −b /(2a) − j (|d |1/2)/(2a) The fact that these are complex conjugates is obscure because the expressions are messy. You might ask. We can be sure that q ≠ 0 because d ≠ 0 and a ≠ 0. We can use the “abbreviated discriminant” version of the quadratic formula for complex roots: x = [−b ± j (|d |1/2)] / (2a) Applying the right-hand distributive rule for addition and subtraction “in reverse” to the right side.

the second and third terms of the polynomial add up to 0. getting x + j5 = 0 This suggests that the binomial factor form of the quadratic is (x − j 5)(x + j 5) = 0 The left side of this equation is 0 if and only if x = j 5 or x = −j 5.Imaginary Roots in Factors
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Imaginary Roots in Factors
Now that we’ve found a way to solve quadratics that have real coefficients. in which the roots are pure imaginary and additive inverses:
3x 2 + 75 = 0 We simplified this equation in the second version of the solution when we divided both sides by 3. Knowing these roots.
A specific case Consider again the quadratic equation we solved earlier. let’s see what the factors of such equations look like. we can rewrite the first term as a sum: [x + (−j 5)](x + j 5) = 0 Applying the product of sums rule gives us x 2 + x j 5 + (−j 5x) + (−j 5)( j 5) = 0 Using the commutative law for multiplication in the third and fourth terms. To avoid confusion with the signs. We’ll start with situations where the coefficient of x is 0. so the roots are pure imaginary. Also. If we take that equation and subtract j 5 from each side. We can therefore simplify to get x 2 + 25 = 0
. obtaining x 2 + 25 = 0 We found that the roots are j 5 and −j 5. In one case. We can add j 5 to each side of that. it’s reasonable to think that we should be able to figure out what this equation looks like in binomial factor form. a real constant. Let’s multiply it out and see what we get. we obtain x 2 + x j 5 + (−x j 5) + (−j × j ) × 25 = 0 Note that −j × j = 1. and a negative real discriminant. we have x = j 5. we have x = −j 5. we get x − j5 = 0 In the other case.

which we know is not 0 because we’ve stated that it’s positive. we get ax 2 = −c Dividing through by a. the coefficient of x is 0.388 Quadratic Equations with Complex Roots
Multiplying through be 3 gives us 3x 2 + 75 = 0 That’s the original equation. we know that the ratio c /a is positive as well. x = −j [(c /a)1/2]
. x = j [(c /a)1/2] If we subtract j [(c /a)1/2] from each side. we obtain x 2 = −c /a which can be rewritten as x 2 = −1(c /a) Because a and c are both positive. and the stand-alone constant is a positive real number c. We can take the square root of both sides of the above equation. In one case.
The general case Consider a general quadratic in which the coefficient of x 2 is a positive real number a. Then we have
ax 2 + c = 0 Subtracting c from each side. we get x − j [(c /a)1/2] = 0 In the other case. Its positive and negative square roots are therefore both real numbers. we can get the binomial factor form of the original quadratic. getting x = ±[−1(c /a)]1/2 = ±(−1)1/2 [±(c /a)1/2] = ±j [(c /a)1/2] = j [(c /a)1/2] or −j [(c /a)1/2] Knowing these roots.

we had better multiply out the left side. and we’ll have the case where a > 0 and c > 0. we can expand to get x 2 + x j [(c /a)1/2] + (−j )[(c /a)1/2]x + {−j [(c /a)1/2]}{j [(c /a)1/2]} = 0 Taking advantage of the commutative law for multiplication in the third and fourth terms. so we can simplify this to x 2 + c /a = 0 Multiplying through by a. “What happens if they are both negative?” The answer is simple. then we can multiply the equation through by −1. we get ax 2 + c = 0 which is the original quadratic. getting (−1) × (−2x 2 − 79) = −1 × 0
. For example. we get x + j [(c /a)1/2] = 0 The binomial factor form of the original quadratic is therefore {x − j [(c /a)1/2]}{x + j [(c /a)1/2]} = 0 To be sure this is the correct equation in binomial factor form.Imaginary Roots in Factors
389
When we add j [(c /a)1/2] to each side. (that is. If a < 0 and c < 0. let’s rewrite the first term as a sum: {x + (−j )[(c /a)1/2]}{x + j [(c /a)1/2]} = 0 Using the product of sums rule. You might ask. we can rewrite it as x 2 + x j [(c /a)1/2] + {−x j [(c /a)1/2]} + (−j × j )(c /a) = 0 The second and third terms are additive inverses and −j × j = 1. to be sure we haven’t made any mistakes!). if we see −2x 2 − 79 = 0 we can multiply each side by −1.
Are you confused?
We’ve seen what happens in a quadratic when the coefficient of x 2 and the stand-alone constant are positive while the coefficient of x is equal to 0. To avoid confusion with the signs.

That means −a > 0. That means −c > 0. we know that −c /a > 0. We can subtract c from each side. we obtain x 2 = −c /a Because −c > 0 and a > 0.390 Quadratic Equations with Complex Roots
which simplifies to 2x 2 + 79 = 0
Here’s a challenge!
Investigate what happens in the general case if a is positive and c is negative in the quadratic equation ax 2 + c = 0
Solution
We have a > 0 and c < 0. We can take the positive-negative square root of both sides to get x = ±(−c /a)1/2 Stated separately. and are additive inverses. getting ax 2 = −c Multiplying through by −1 gives us (−a)x 2 = c
. getting ax 2 = −c Dividing through by a.
Here’s another challenge!
Investigate what happens in the general case if a is negative and c is positive in the quadratic equation ax 2 + c = 0
Solution
We have a < 0 and c > 0. Let’s rewrite the above equation as ax 2 − (−c) = 0 We can add −c to each side. the roots are x = (−c /a)1/2 or x = −[(−c /a)1/2] These are both real numbers.

we know that c /(−a) > 0.
Conjugate Roots in Factors
We’ve seen the binomial factor forms of quadratics where the roots are pure real or pure imaginary. Then we can generate the binomial factor form of the quadratic from those roots.
An example We can use the quadratic formula to find the roots of a quadratic equation that has complex conjugate roots. The discriminant is d = b 2 − 4ac = 62 − 4 × 5 × 5 = 36 − 100 = −64 The square root of the discriminant is pure imaginary: ±(d 1/2) = ±[(−8)1/2] = ±j 8
. Now let’s look at the factors when the roots are complex conjugates. but are not pure imaginary numbers. and c = 5 in the polynomial standard form. b = 6. Let’s try this:
5x 2 + 6x + 5 = 0 We have a = 5.Conjugate Roots in Factors
Dividing through by (−a). we obtain x 2 = c /(−a)
391
Because c > 0 and −a > 0. Both roots are real. the roots are x = (−c /a)1/2 or x = −[(−c /a)1/2] These are the same roots we get if a > 0 and c < 0. and they are additive inverses of each other. We can take the positive-negative square root of both sides to get x = ±[c /(−a)]1/2 which can be rewritten as x = ±(−c /a)1/2 Stated separately.

It can be extrapolated to all real. remember the rule you derived when you solved Practice Exercise 10 at the end of Chap. Also suppose that the discriminant. verifying that it’s equivalent to the original general quadratic in polynomial standard form. imaginary. 9. review it now. where d = b 2 − 4ac Write the general quadratic equation in factored form. If you’ve forgotten how that rule works.
Here’s a challenge!
Consider a general quadratic equation in polynomial standard form: ax 2 + bx + c = 0 Suppose that b is not equal to 0. is negative.
Solution
We can simplify things if we temporarily rename two of the terms in the binomial.Conjugate Roots in Factors
393
Are you confused?
If you wonder how the two trinomials in the above situation can be multiplied out. Let p = −b /(2a) and q = (|d |1/2)/(2a)
. d. and complex numbers. We’ll give them their “legitimate” names back later.
Solution
Let’s begin with the quadratic formula we obtained earlier in this chapter for cases of this sort: x = [−b ± j (|d |1/2)] / (2a) We can break the numerator apart and write this as the sum of two fractions with a common denominator: x = −b /(2a) ± j (|d |1/2)/(2a) The roots can be expressed separately like this: x = −b /(2a) + j (|d |1/2)/(2a) or x = −b /(2a) − j (|d |1/2)/(2a) This suggests that the factored form of the equation ought to be {x − [−b /(2a) + j (|d |1/2)/(2a)]}{x − [−b /(2a) − j (|d |1/2)/(2a)]} = 0
Here’s another challenge!
Multiply out the factored equation we’ve just obtained.

Don’t hurry! You’ll find worked-out answers in App. 6. Suppose we want the roots to be x = j 7 or x = −j 3 Write down the binomial factor form of a quadratic with these two roots. 10. Convert the equation from the solution to Prob. Plug the roots from the solution to Prob. Convert the equation given in Prob. However. 2. 8. What are the roots of the following quadratic equation? What is the solution set X ? (x − j 7)(x + j 7) = 0 2. 5. Convert the following quadratic into polynomial standard form: (j 2x + 2 + j 3)(−j 5x + 4 − j 5) = 0
. 9. you should be able to solve them using the rules and techniques you’ve been taught so far. C. We haven’t dealt with quadratic equations in which the roots are pure imaginary but are not additive inverses. Use the quadratic formula to find the roots of the polynomial equation as it is expressed in the solution to Prob. What are the roots of the following quadratic equation? What is the solution set X ? (x + 2 + j 3)(x − 2 − j 3) = 0 8. 3. we can “manufacture” such an equation. Use the quadratic formula to find the roots of the polynomial equation as it is expressed in the solution to Prob. The solutions in the appendix may not represent the only way a problem can be figured out. Convert the equation given in Prob. and be careful with the signs and j operators! 1. 5. 4. 7 into the polynomial standard equation from the solution to Prob. 7 into polynomial standard form. 4 into polynomial standard form.
395
Practice Exercises
This is an open-book quiz. Nevertheless. 1 into polynomial standard form. we obtain ax 2 + bx + c = 0 which is the original general quadratic in polynomial standard form. showing that those roots actually work. Be patient. by all means try it! Note: Some of these problems take you a little beyond the material covered directly in the text of this chapter. If you think you can solve a particular problem in a quicker or better way than you see there. You may (and should) refer to the text as you solve these problems.Practice Exercises
and further to x 2 + (b /a)x + c /a = 0 Multiplying through by a. 7.

Two Real Zeros y
397
x
Figure 24-1 The graphs of quadratic functions with
real coefficients and a real constant are always parabolas. Imagine that its quadratic function is
y = ax 2 + bx + c The x-intercepts are r and s. At the points on the x axis where x = r and x = s. They represent the roots of the equation ax 2 + bx + c = 0 so they can be found using the quadratic formula.
Parabola opens upward Figure 24-2 shows a generic graph of a quadratic function of x with two real zeros. If r is the smaller of the two zeros and s is the larger. the parabola crosses. This parabola opens upward. then r = [−b − (b2 − 4ac)1/2] / (2a) and s = [−b + (b2 − 4ac)1/2] / (2a)
. and they always pass the “vertical-line test” for a function. which we call r and s.

” This point is called the absolute minimum of the graph.398 Graphs of Quadratic Functions y
x x=r y=0 xmin = (r + s) / 2 y<0 Absolute minimum x=s y=0
Figure 24-2 Graph of a quadratic function with
two real zeros when the coefficient of x 2 is positive. The parabola opens upward. crosses the x axis twice. then
r = [−b + (b2 − 4ac)1/2] / (2a) and s = [−b − (b2 − 4ac)1/2] / (2a) Note the subtle difference between these two equations and those for the case where the parabola opens upward! Because of this. again called r and s. In a few moments. the values of r and s are reversed. as compared to their
. But this parabola opens downward.
Whenever a parabola opens upward. If r is the smaller of the two zeros and s is the larger. and has an absolute minimum with y < 0.
Parabola opens downward Figure 24-3 shows a generic graph of a quadratic function of x with two real zeros. it has a single point at which it “bottoms out. we’ll discover how this point can be located.

Two Real Zeros y xmax = (r + s) / 2 y>0 Absolute maximum x=r y=0 x=s y=0 x
399
Figure 24-3 Graph of a quadratic function with two
real zeros when the coefficient of x 2 is negative. “absolute extreme value.
Alternative function notation Sometimes a quadratic function is given a name such as f. The parabola opens downward. the extremum may also be called the vertex. Sometimes the term extremum is used in reference to an absolute maximum or an absolute minimum in the graph of a function. we might write
f (x) = x 2 + 2x + 1 instead of y = x 2 + 2x + 1
.” This point is called the absolute maximum. When a parabola opens downward. For example. it has a single point at which it “peaks. It means. crosses the x axis twice.
values when the parabola opens upward. and then its value is denoted by that name with the independent variable in parentheses afterward.” In a parabola. as you might guess. and has an absolute maximum with y > 0.

and draw an approximate curve through these three points that represents the graph of the function. It’s the arithmetic mean. or average. If a > 0 in the polynomial. we can plug the x-value into the function once we’ve found it. Let’s call its x-value xmin. and its graph is not a parabola. and we have an absolute minimum somewhere. then we don’t have a quadratic function at all. then again xmax = (r + s)/2 To find the y-value of the absolute minimum or maximum. you might ask. When we have plotted the two x-intercepts along with the extremum. if a > 0 in the general form of the function). If the coefficient of x 2 in the polynomial is larger than 0 (that is. plot the zeros and the absolute minimum or maximum. then the parabola opens upward. Then find the two real zeros. of the zeros r and s. We’re lucky here because the polynomial equation factors into (x − 1)(x − 2) = 0
. If we call its x-value xmax. we can draw a fair approximation of the parabola representing the quadratic function.
Here’s a challenge!
Consider the following quadratic function: y = x 2 − 3x + 2 Determine whether the parabola for this function opens upward or downward.400 Graphs of Quadratic Functions
Are you confused?
By now. then the parabola opens downward. find the x-value of the extremum. then the parabola opens upward. If a < 0. Then
xmin = (r + s)/2 If a < 0. it’s easy to find the x-value of the absolute minimum or maximum of its graph. so we know that the parabola opens upward.
Solution
The coefficient of x 2 is positive. and it has an absolute maximum.
Finding the absolute minimum or maximum When we know the x-intercepts of a quadratic function with two real zeros. Then we can plot the point on a coordinate grid. “How can we can tell whether the parabola for a particular quadratic function opens upward or downward?” This is easy to figure out. r and s. Finally. A little arithmetic will give us the y-value. After that. then the parabola opens downward. If a = 0. Then determine its y-value.

“tangent” means that the curve just brushes against the axis. At the point (r. we can fill out the right-hand side without having to find another distant point there.
Parabola opens upward Figure 24-5 shows a generic graph of a quadratic function of x with one real zero r. The root of the equation ax 2 + bx + c = 0 can be found using the quadratic formula.
Here’s a trick!
The graph of a quadratic function is bilaterally symmetric with respect to a vertical line passing through the vertex. touching it at a single point. this fact can be useful. That means the right-hand side of the curve is an exact “mirror image” of the left-hand side. 0). Because there is only one root. the parabola is tangent to the x axis. When we plug the value 0 in for x.
One Real Zero
When a quadratic equation has one real root. Here. −1/4). When we graph such a function. It is also shown in Fig. 2) is on the curve.402 Graphs of Quadratic Functions
to get a clear picture of the parabola based on their locations. In the challenge we just finished. (1. we get y = x 2 − 3x + 2 = 02 − 3 × 0 + 2 =0−0+2 =2 This tells us that (0. But we can find the y-intercept to help us draw the curve. the discriminant in the quadratic formula must be 0. 2). 24-4. If the quadratic function is
y = ax 2 + bx + c then a > 0 because the parabola opens upward. and (0. Once we’ve drawn an approximation of the left-hand side of the curve using the points (3/2. it has a single point in common with the independent-variable axis. That means b2 − 4ac = 0 Therefore r = [−b ± (b2 − 4ac)1/2] / (2a) = −b/(2a)
. 0). its quadratic function has one real zero.

is tangent to the x axis at a single point.One Real Zero y x=q y=k Line y = k where k > 0
403
x=p y=k x x=r y=0 Absolute minimum
Figure 24-5 Graph of a quadratic function with one
real zero when the coefficient of x 2 is positive. And just as before. The root of the equation ax 2 + bx + c = 0 can again be found using the quadratic formula. 0) on the parabola is tangent to the x axis. intersects the parabola twice. the point (r. so r = −b /(2a)
. the discriminant in the quadratic formula is equal to 0. and has an absolute minimum with y = 0. where k > 0. 0). The parabola seems to “sit upon” the x axis. as before.
The absolute minimum is at the point (r. If the quadratic function is
y = ax 2 + bx + c then a < 0 because the parabola opens downward. Also as before. A line with the equation y = k. Let’s call it r. The parabola opens upward.
Parabola opens downward Figure 24-6 illustrates the graph of another generic quadratic function with one real zero.

A line with the equation y = k. 24-5 and 24-6. We can find the x-values of those points by letting y = k in the quadratic function. If the parabola opens upward.404 Graphs of Quadratic Functions y
x=r y=0 Absolute maximum x x=q y=k
x=p y=k
Line y = k where k < 0
Figure 24-6 Graph of a quadratic function with one
real zero when the coefficient of x 2 is negative. we should choose a negative number for k. intersects the parabola twice. If the parabola opens downward.
Are you confused?
In Figs. once we’ve found the zero point (r. we should choose a positive number for k. 0). is tangent to the x axis at a single point. 0) and have figured out whether the parabola opens upward or downward. you’ll notice horizontal lines with the equation y = k. Then we’ll know that the line y = k must intersect the parabola at two points. and then “cooking up” a new quadratic equation out of that: ax 2 + bx + c = k
. You might ask. “Why are the lines there? How do we find the points where the lines intersect the parabolas? Why do we need the points?” The answer: “Curve construction!” The lines allow us to find points that help us draw approximations of the graphs.
The absolute maximum is at the point (r. The parabola opens downward. The graph looks as if the parabola “hangs from” the x axis. and has an absolute maximum with y = 0. where k < 0.

we get x = [−b ± (b2 − 4ac)1/2] / (2a) = {−12 ± [122 − 4 × (−4) × (−9)]1/2} / [2 × (−4)] = [−12 ± (144 − 144)1/2] / (−8) = (−12 ± 02) / (−8) = −12/(−8) = 3/2 The equation has only one root. then we have r = 3/2. and we’ll have an easy time drawing an approximation of the parabola. we learn the coordinates of three points that lie on the parabola. k). That means the parabola opens downward. draw an approximation of the curve.One Real Zero
This can be rearranged to get ax 2 + bx + (c − k) = 0
405
which we can solve with the quadratic formula. getting a horizontal line that crosses the parabola twice. and the parabola “hangs from the x axis. b = 12. so the function has only one zero. The roots of this quadratic. Those points are (p. are the x-values of the two points where the parabola intersects the horizontal line y = k. and (r. Determine whether the parabola for this function opens upward or downward. When we plug these into the quadratic formula. We also know that this point represents the extremum. If we call it r.
Solution
We can find the zero or zeros of the function by applying the quadratic formula to the equation −4x 2 + 12x − 9 = 0 In the general polynomial equation ax 2 + bx + c = 0 we have a = −4. k). so (3/2. and c = −9. The value of a. If we’ve chosen the value of k wisely (made a lucky guess). the coefficient of x 2 in the polynomial. 24-7. (q. 0). Finally. and we know that the point (3/2. as shown in Fig. we’ll have three well-separated points. 0) is on the parabola. 0) is the absolute maximum. Then find two more points so the curve can be drawn. By doing all this. Find the x-value and the y-value of the extremum.” We can choose a negative number k and set it equal to y. Let’s try −5 for k. is negative. which we’ll call p and q.
Here’s a challenge!
Look at the following quadratic function: y = −4x 2 + 12x − 9 Find the zero or zeros. The equation of the line is y = −5.
.

−5) and (21/8. but the curve lies entirely on one side of the independentvariable axis. With these three points put down on our graph paper. ymin. and we can pick
. because the parabola opens upward.
No Real Zeros
When a quadratic equation has no real roots and its associated quadratic function has no real zeros. We’ve already determined that (3/2. The curve has an absolute minimum that lies in the first or second quadrant. −5). The parabola does not cross the x axis anywhere. we get roughly 8. when we draw the curve on paper. we can find the coordinates of two other points. If we call that value xmin. substituting 9 in for 801/2. “Where do we start locating points on the curve in situations like this?” The task doesn’t look easy at first.944. It’s reasonable to wonder. then xmin = −b /(2a) The y-value of the absolute minimum point. We can plot the points as (3/8.No Real Zeros
407
Our goal is to find the coordinates of points so we can draw an approximation of the parabola. When we use a calculator to find 801/2. and that it represents the absolute maximum. is whatever we get when we plug xmin into the function: ymin = axmin2 + bxmin + c Once we’ve found the coordinates of the absolute minimum. we can fill in an approximation of the parabola. We can’t get perfection with pencil and paper. as follows: x = (−12 ± 9) / (−8) = −3/(−8) or −21/(−8) = 3/8 or 21/8 These represent the approximate x-values of the points where the horizontal line intersects the parabola in Fig. 24-7. If the quadratic function is
y = ax 2 + bx + c then a > 0 in this example. That’s so close to 9 that it won’t make a difference. So let’s continue with our arithmetic. If we can get within 1/10 of a unit. The y-values are exactly equal to −5 for both points. 0) is on the curve. its graph is a parabola. if we call it 9. but we can find the x-value of the absolute minimum point. so we don’t need exact values for the coordinates. that ought to be good enough. We can pick a number p somewhat smaller than xmin.
Parabola opens upward Figure 24-8 is a generic graph of a quadratic function of x with no real zeros.

408 Graphs of Quadratic Functions y x=p y = f (p)
x=q y = f (q)
x xmin = –b/(2a) y>0 Absolute minimum
Figure 24-8 Graph of a quadratic function with no real
zeros when the coefficient of x 2 is positive. we have to go through the arithmetic twice. does not cross the x axis. By choosing the y-value first. and figure the y-value by plugging into the function? Isn’t that better than picking a y-value and then grinding through the quadratic formula to get two x-values that are likely to come out irrational?” You tell me! By choosing x-values first.” you might ask. has an absolute minimum with an x-value equal to −b/(2a). but we only have to do it once. we can plug them into the function for x and find two more points on the curve. The numbers p and q don’t have to be equally smaller and larger than xmin. Either method will work fine.
Are you astute?
Do you wonder why we didn’t use the “x-value first” point-finding strategy earlier in this chapter for quadratic functions with one or two real zeros? “Isn’t it easier. Once we’ve chosen the numbers p and q. The parabola opens upward. the arithmetic can be a little rough. but it’s usually simple.
another number q somewhat larger than xmin.
. and has a positive y-value. “to choose a clean integer x-value for a point. although we should try to get close to that ideal. It’s your choice. Then the parabola is easy to draw. We can choose integers for these numbers to make the arithmetic easy.

The parabola opens downward. is ymax = axmax2 + bxmax + c Once we’ve found the coordinates of the vertex.No Real Zeros
409
Parabola opens downward Figure 24-9 is another generic graph of a quadratic function of x with no real zeros. If the function is
y = ax 2 + bx + c then a < 0 because the parabola opens downward. the parabola does not cross the x axis. The curve has an absolute maximum in the third or fourth quadrant. is xmax = −b/(2a) The y-value of the absolute maximum point. The x-value of the absolute maximum point. we can find the coordinates of two other points. xmax. Then the parabola is easy to draw. ymax. Again.
y xmax = –b/(2a) y<0 Absolute maximum
x
x=p y = f (p)
x=q y = f (q)
Figure 24-9 Graph of a quadratic function with no real
zeros when the coefficient of x 2 is negative. and has a negative y-value. does not cross the x axis. We can pick a number p smaller than xmax. has an absolute maximum with an x-value equal to −b/(2a). We can then plug p and q into the function for x to find two more points on the curve. and we can pick another number q larger than xmax.
.

” and it’s always the vertex in a quadratic function with real coefficients and a real constant. It’s also true for quadratic functions that have one or two real zeros. Differential calculus gives us a formula for the slope m of a quadratic function at any point based on the x-value at that point: m = 2ax + b If we set m = 0 to make the slope horizontal. is always horizontal. This holds true whether the parabola opens upward or downward. we get 0 = 2ax + b That’s a first-degree equation in x. find the x-value and the y-value of the extremum. That means it has an absolute minimum. Then locate two other points on the curve. It involves finding the point on the parabola where a line tangent to the curve has a slope of 0. That’s the point where the curve is “locally horizontal. and draw an approximation of the parabola representing the function. we get x = −b/(2a)
Here’s a challenge!
Consider the following quadratic function with no real zeros: y = 2x 2 + 4x + 3 Find out whether the parabola for this function opens upward or downward. When we solve it. Do you wonder how we know this? I haven’t proven it or even derived it in a general way.410 Graphs of Quadratic Functions
Are you confused?
We’ve made a substantial claim: The x-value of the extremum for a quadratic function y = ax 2 + bx + c is equal to −b/(2a) when there are no real zeros. because the coefficient of x 2 is positive. you’ll see that a straight line drawn tangent to the curve. After that. Remember again the general polynomial form for a quadratic function: y = ax 2 + bx + c The x-value at the absolute minimum point is xmin = −b/(2a) = −4 / (2 × 2) = (−4)/4 = −1
. and passing through the vertex.
Solution
This parabola opens upward. This fact comes out of a maneuver that requires differential calculus. but the approach can be explained qualitatively. We won’t get into calculus in this book. If you look at all the graphs in this chapter.

. What are the real zeros. Draw an approximate graph of the function stated in Prob. 1? 3. The solutions in the appendix may not represent the only way a problem can be figured out. and (1. What are the coordinates of the vertex point on the parabola representing the function stated in Prob. You may (and should) refer to the text as you solve these problems. if any. 9). Don’t hurry! You’ll find worked-out answers in App. Figure 24-10 shows these points along with an approximation of the parabola passing through them. Consider the following quadratic function: y = −2x 2 + 2x − 5 Does the parabola representing the graph of this function open upward or downward? 8. 1). 7? 10. 1? 4. If you think you can solve a particular problem in a quicker or better way than you see there. Draw an approximate graph of the function stated in Prob. What are the coordinates of the point representing the extremum of the function stated in Prob. What are the coordinates of the point representing the vertex of the parabola for the function stated in Prob. if any. We now have three points on the curve: (−3. 7? 9. 1. of the function stated in Prob. (−1. 9). by all means try it! 1. 9). 5? 7. C. 5. Examine this quadratic function: y = (x − 3)(4x − 1) Does the parabola representing the graph of this function open upward or downward? 2.412 Graphs of Quadratic Functions
y = 2x 2 + 4x + 3 = 2 × 12 + 4 × 1 + 3 =2+4+3 =9 The third point is therefore (1.
Practice Exercises
This is an open-book quiz. How can we know this without plotting any points? y = 7x 2 + 5x + 2 6. 7. of the function stated in Prob. The graph of the following quadratic function lies entirely above the x axis. What are the real zeros.

the coefficient of x in the binomial is 2. or difficult. we get 8x 3 − 36x 2 + 54x − 27 = 0
What’s the real root? A single-variable equation in binomial-cubed form has one real root. the coefficient of x in the binomial is 1 and the stand-alone constant is 3.
. consolidate the terms for x 2 and x. and then dividing through by a. and the constant is −3. Therefore.
Multiplying out Now look at the last equation shown in the above set of three. challenging. the stand-alone constant is 0. we can rewrite it as
(2x − 3)(2x − 3)(2x − 3) = 0 Then we can multiply the second two factors using the product of sums rule. and pay close attention to the signs. In general. and complicated cubics can sometimes look simple.) The result is (2x − 3)(4x 2 − 12x + 9) = 0 If we multiply again using the expanded product of sums rule. In the third case. We can multiply it out so the left side becomes a polynomial. (Remember that subtraction is the same as the addition of a negative. assuming the coefficient and the constant are both real numbers. because a ≠ 0. We often cannot know by glancing at a third-degree equation whether solving it will be easy. we can write (x + 0)3 = 0 In the second case. That root can be found by taking away the exponent from the binomial. and then solving the first-degree equation that remains.414 Cubic Equations in Real Numbers
In the first case. We can get away with division by a. the single real root to the cubic is x = −b /a
Are you confused?
Simple cubics can sometimes look complicated. If we want to be formal. First. if we have the third-degree equation
(ax + b)3 = 0 then the real root is the solution to ax + b = 0 That solution is obtained by subtracting b from both sides. The following example can illustrate.

Three Binomial Factors
415
Here’s a challenge!
Solve the following cubic equation. Two of them might be identical. 9. so this equation looks difficult! But for the moment. and discover that the real-number root is an integer: (31/2x − 121/2)3 = 0
Solution
Remember that the 1/2 power of a number is the positive square root. we can simplify this root and then reduce it to an integer: x = 121/2 / 31/2 = (12/3)1/2 = 41/2 =2 We should check to be sure that this root really works in the original equation! Let’s plug it in and find out: (31/2x − 121/2)3 = 0 (31/2 × 2 − 121/2)3 = 0 (31/2 × 41/2 − 121/2)3 = 0 Using the power of product rule from Chap.
. Therefore x = −b /a = −[−(121/2)] / 31/2 = 121/2 / 31/2 Using the power of quotient rule from Chap. It’s a legitimate cubic equation in x. because the 1/2 powers involve only the coefficient and the constant. those factors are rarely all the same. not the variable x itself. but often all three are different. We can solve it using the above general formula. Both the coefficient and the stand-alone constant are irrational. we can simplify this to [(3 × 4)1/2 − 121/2]3 = 0 (121/2 − 121/2)3 = 0 03 = 0 0=0
Three Binomial Factors
When a cubic equation can be expressed as three binomial factors. The above equation is therefore not ambiguous. 9. let’s forget about that. Cubics of this sort are in binomial-factor form. We have a = 31/2 and b = −(121/2).

so the multiplied-out equation contains a nonzero multiple of x 3. a2. Here are some examples of binomial-factor cubics: (x − 1)(x + 2)(x − 3) = 0 (3x + 2)(5x + 6)(−7x − 1) = 0 x (x + 6)(x + 8) = 0 x 2(−4x − 1) = 0 The third and fourth of these equations have one and two stand-alone constants equal to 0. these equations are (x + 0)(x + 6)(x + 8) = 0 and (x + 0)(x + 0)(−4x − 1) = 0
Multiplying out We’re fortunate if we can reduce a cubic to three binomial factors. The three stand-alone constants are b1.416 Cubic Equations in Real Numbers
Binomial-factor form Here’s the general form of a cubic broken down into three binomial factors. In “unabridged” binomial-factor form. and a3. respectively. and b3. Start with this: x (x + 6)(x + 8) = 0 If we multiply the first factor by the second. b2. assuming that the equation can be expressed that way with real coefficients and real constants:
(a1x + b1)(a2x + b2)(a3x + b3) = 0 The three coefficients are a1. Consider the equation
x 3 − 2x 2 − 5x + 6 = 0 This doesn’t advertise that it can be factored into a product of binomials! But try multiplying this out: (x − 1)(x + 2)(x − 3) = 0 Let’s work an example backward. we obtain (x 2 + 6x)(x + 8) = 0 Multiplying these two factors out gives us x 3 + 14x 2 + 48x = 0
. They must all be nonzero.

Even if we couldn’t factor the trinomial. we get x = −6/5. “Why not divide the following equation through by x in an attempt to simplify and solve it?” x 3 + 14x 2 + 48x = 0
. and then test these roots in the resulting equation. we do the same thing with the second binomial factor. We would notice that the stand-alone constants must add up to 14 and multiply to 48. we solve the equation −7x − 1 = 0 Adding 1 to each side and then dividing through by −7. as the next example will show. we get x = 1/(−7) = −1/7. we get x = −2/3.
Are you confused?
You might wonder. The real roots of the original cubic are therefore x = −2/3 or x = −6/5 or x = −1/7 and the solution set X is {−2/3. That equation is 3x + 2 = 0 Subtracting 2 from each side and then dividing through by 3. Later. −1/7}.
What are the real roots? To see how we can solve a binomial-factor cubic with real coefficients and a real constant. it wouldn’t take us long to guess that they are 6 and 8. We could factor x out. we could attack it with the quadratic formula. we solve the equation created by setting the first binomial factor equal to 0. In Practice Exercises 3 and 4 at the end of this chapter. To find the third real root. getting x (x 2 + 14x + 48) = 0 We could try to break up the trinomial part of this equation into binomial factors. Once we had the three binomial factors of the cubic. we’d find them. we’ll see an example of that tactic. finding the real roots would be easy.Three Binomial Factors
417
Imagine that we were presented with the above equation for the first time. We have 5x + 6 = 0 Subtracting 6 from each side and then dividing through by 5. you’ll multiply out the factors of the original cubic. and we had never seen it in factored form. let’s find the real roots of
(3x + 2)(5x + 6)(−7x − 1) = 0 To find the first real root. With a few trials and errors. To find the second real root. −6/5.

we take the first-degree equation x+8=0 which resolves to x = −8. Suppose that we could look at the above equation and see the binomial factor version immediately: x (x + 6)(x + 8) = 0 To find the first root. −8}.418 Cubic Equations in Real Numbers
That’s a reasonable question. To find the third root. because it’s incomplete. Dividing through by x can “knock the equation down” from third degree to second degree—or so we might be tempted to believe. Let’s try it and see what happens! We get (x 3 + 14x 2 + 48x ) / x = 0/x which apparently simplifies to x 3/x + 14x 2/x + 48x/x = 0 and further to x 2 + 14x + 48 = 0 This is a quadratic that can be factored into (x + 6)(x + 8) = 0 The roots x = −6 or x = −8 come out of this process. That’s easy. We conclude that the solution set X for the original cubic must be {−6. The real roots of the cubic are therefore x = 0 or x = −6 or x = −8
. we can take literally the simple first-degree equation x=0 To find the second root.
Here’s a challenge!
Solve the following cubic equation in a way that works properly: x 3 + 14x 2 + 48x = 0
Solution
We’ve seen this equation in its binomial factor form. Let’s find out how to get it right. we take the first-degree equation x+6=0 which resolves to x = −6. isn’t it? Not so fast! It’s also wrong.

the equation is in binomial-trinomial form. if any such roots exist.Binomial Times Trinomial
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and the solution set X is {0. any or all of which can equal 0. b2. the coefficient of x is 0 in the trinomial. −6. Here’s a good one. with plenty of sign changes to make it interesting:
(−4x − 3)(−7x 2 + 6x − 13) = 0 Using the expanded product of sums rule. we get 28x 3 − 3x 2 + 34x + 39 = 0
. A new root shows up this time: x = 0! The fact that one of the roots is 0 caused us to inadvertently divide by 0 when we divided the equation through by x.
Multiplying out Let’s take a specific example of a cubic in binomial-trinomial form and multiply it out. The technique shown in this section will also reveal the complex-number roots of a cubic equation. and c are real numbers. Also suppose that b1. the stand-alone constant is 0 in the binomial. This “blinded” us to the existence of that root. The binomial-trinomial form of a cubic equation in the variable x can be written as follows:
(a1x + b1)(a2x2 + b2x + c) = 0 Here are some examples of cubics in the binomial-trinomial form: (−4x − 3)(−7x 2 + 6x − 13) = 0 (3x + 5)(16x 2 − 56x + 49) = 0 (3x)(4x 2 − 7x − 10) = 0 (−21x + 2)(3x 2 − 14) = 0 In the third case above.
Binomial Times Trinomial
When a cubic can be expressed as a binomial multiplied by a trinomial. (Actually. −8}. I’ve never seen that expression used in other texts. But it’s easy to remember. we obtain 28x 3 − 24x 2 + 52x + 21x 2 − 18x + 39 = 0 Consolidating the terms for x 2 and x.
Binomial-trinomial form Suppose that a1 and a2 are nonzero real numbers. In the fourth case. don’t you think?) A cubic in this form is not particularly difficult to solve for real roots.

as long as all the coefficients and constants are real numbers.) Expressed for the above general equation. and that equation always has a real solution. or none at all. we get a2x 2 + b2x + c = 0 We can find the real roots of this equation. or only one real root. (3x + 5)(16x 2 − 56x + 49) = 0
Solution
First. we can construct a first-degree equation by setting the binomial factor equal to 0. That’s because a first-degree equation can always be created from the binomial factor. that would be
a1x + b1 = 0 which solves to x = −b1/a1 This will always give us one real root for the cubic. the quadratic formula will produce them. First. along with two others that are complex. In the general form above.420 Cubic Equations in Real Numbers
What are the real roots? The process of finding the real roots of a cubic in the binomial-trinomial form is straightforward. How about cubics? You’ve already seen an example of a cubic with three real roots. That gives us 3x + 5 = 0
. if the root or roots are complex but not real. In the general form shown above. We can take this statement further: A cubic equation. “How about no real roots?” The answer: Any cubic in the binomialtrinomial form with real coefficients and a real constant always has at least one real root. Then you’ll discover that a cubic in the binomial-trinomial form with real coefficients and a real constant can have only one real root. (I like to use the quadratic formula. the quadratic formula is x = [−b2 ± (b22 − 4a2c)1/2] / (2a2)
Are you confused?
You’ve learned that a quadratic equation can have two different real roots. if any exist.
Here’s a challenge!
Find the real roots of the following cubic equation using the method described in this section. we “manufacture” a first-degree equation from the binomial. because it always works! Also. setting it equal to 0. no matter what the form. has at least one real root if all the coefficients and constants are real numbers. and state the real solution set X. using techniques we’ve already learned for solving quadratics. “Okay. You’re about to see that a cubic equation in the binomial-trinomial form with real coefficients and a real constant can have two real roots. Then you ask.” you say. After that. obtaining a quadratic equation. we set the trinomial equal to 0.

The original cubic therefore has only one real root. The quadratic we obtain by setting the trinomial factor equal to 0 has no real roots (although it has two complex roots). you’ll check these roots. which is equal to 11/2. we get x = −5/3.136)1/2] / 32 = 56/32 = 7/4 In this case. Then we apply the quadratic formula to the trinomial factor.
. b2 = −56. The original cubic therefore has two real roots: x = −5/3 or x = 7/4 and the real solution set X is {−5/3. then x = [−b2 ± (b22 − 4a2c)1/2] / (2a2) = {56 ± [(−56)2 − 4 × 16 × 49]1/2} / (2 × 16) = [56 ± (3. and c = 49.
Here’s another challenge!
Find the real roots of the following cubic equation using the approach described in this section. we get the solution x = −11/(−2). and c = 5. and state the real solution set X.136 − 3. so X = {11/2}. Next.” is to create a first-degree equation by setting the binomial factor equal to 0. because the discriminant is negative. That gives us −2x + 11 = 0 When we subtract 11 from each side and then divide the entire equation through by −2. (−2x + 11)(2x 2 − 2x + 5) = 0
Solution
Our first step. as in the previous “challenge. x = 11/2. b2 = −2. If we let a2 = 2.Binomial Times Trinomial
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Subtracting 5 from each side and then dividing through by 3. we apply the quadratic formula to the trinomial factor. 7/4}. The root x = 7/4 has multiplicity 2. we have x = [−b2 ± (b22 − 4a2c)1/2] / (2a2) = {2 ± [(−2)2 − 4 × 2 × 5]1/2} / (2 × 2) = [2 ± (4 − 40)1/2] / 4 = [2 ± (−36)1/2] / 4 We can stop right here. 7/4 is the only real root of the quadratic we obtain by setting the trinomial factor equal to 0. The real solution set X contains only that single root. If we let a2 = 16. In Practice Exercise 7 at the end of this chapter.

422 Cubic Equations in Real Numbers
Polynomial Equation of Third Degree
Any single-variable cubic equation can be written so it appears in polynomial standard form. Often. Here’s the general form in the realm of the real numbers:
ax 3 + bx 2 + cx + d = 0 where a. perhaps without having to do any manipulations. a binomial-trinomial expression for the equation must exist.
. as well as the stand-alone constant. If we come across the equation
x3 − 8 = 0 we can see. If we can manage to do that. All of the following equations are cubics in this form: x 3 + 3x 2 + 2x + 5 = 0 −2x 3 + 3x 2 − 4x = 0 5x 3 − 7x 2 − 5 = 0 −7x 3 − 4x 2 = 0 4x 3 + 3x = 0 −7x 3 − 6 = 0 If you set a = 0 in the polynomial standard form of a single-variable cubic equation. can equal 0 in a cubic polynomial. b. then we can find the roots as described in the previous sections. and the right side contains 0 alone. a ≠ 0. But factoring a cubic polynomial is not always easy. we can try to factor it into a product of binomials. but often they are not. this is the form you’ll first see. c. you end up with the polynomial standard form for a single-variable quadratic. the roots may be apparent right away. But if we encounter x 3 + 3x 2 + 2x + 5 = 0 the situation is more challenging. If the coefficients and the constant of the polynomial equation are all real numbers. x = 2.
Polynomial standard form When a cubic equation is in polynomial standard form. the left side of the equals sign contains a third-degree polynomial. the coefficients of x 2 or x. However. They might even turn out to be irrational numbers. but the coefficients and constants might not be integers.
What can we do? When we see a cubic equation in polynomial standard form. that there’s one real root. or into binomial-trinomial form. or even a first-degree equation (depending on the other coefficients). When we see an equation like this. and x is the variable. and d are real numbers.

It seems reasonable to suppose that if we can find the real root k. This rule. Remember what “if and only if ” means in logical terms: the “if-then” reasoning works both ways. any positive-integer degree (called nth-degree equations). If we plug in k for the variable in the equation—no matter what form that equation happens to be in—and work out the arithmetic. we fail Now that we’re armed with plenty of theoretical facts. then k is a real root of the cubic equation. the fifth degree (called quintic equations). we should formalize an important principle. Now imagine that we are faced with a cubic equation in polynomial standard form. in fact. and some bad news. the result will be 0. can be generalized to polynomial equations in the fourth degree (called quartic equations). We can package all this reasoning up into a formal statement called the binomial factor rule or the factor theorem:
• A real number k is a root of a cubic equation in the variable x if and only if (x − k) is a factor of the cubic polynomial. and with 0 as one of the factors. and we can’t figure out how it can be factored. Let’s try this:
x 3 − 7x − 6 = 0 Remember the general form: ax 3 + bx 2 + cx + d = 0
.
The binomial factor rule A little while ago. it’s time to take aim at a real polynomial cubic equation using synthetic division. First. That often works. We can therefore rewrite the above rule as two separate statements: • If a real number k is a root of a cubic equation in the variable x. Finally.
We try. the good news: We can try to factor a binomial out of a polynomial using a process called synthetic division.Polynomial Equation of Third Degree
423
If we see a cubic in polynomial standard form and we find ourselves staring at it. Before we see how synthetic division works. but we’ll probably have to go through the process several times before we find the right binomial. and. the whole expression attains the value 0. then the binomial (x − k) can be factored out of the cubic. which must exist. then (x − k) is a factor of the cubic polynomial. I made a strong claim: Any cubic equation with real coefficients and a real constant has at least one real root. then the value of (x − k) becomes 0. which has been proven as a theorem by mathematicians. there’s some good news. the bad news: That formula is so complicated that most mathematicians would rather try every other option first. some fair news. • If k is a real number and (x − k) is a factor of a cubic polynomial in the variable x. That’s because if we set x equal to k. Now for the fair news: A general formula for solving a cubic equation exists. Suppose we call that real number k. paralyzed with uncertainty.

” we write down the coefficients and the standalone constant in an array along with “wildcard” entries symbolized by pound signs (#). and d = −6. writing the sum in the bottom row:
2 1 1 0 2 2 −7 # # −6 # #
. c = −7. b = 0. like this:
# a # b # # c # # d # #
In our example. starting immediately to the right of the pound sign:
# 1 # 0 # # −7 # # −6 # #
Now the guessing game begins! At first. we have a = 1. we have no clue as to what the real root might be. Let’s try 2 as a “test root. we copy the second number in the top row to the “wildcard” slot directly beneath it in the bottom row:
2 1 1 0 # # −7 # # −6 # #
We multiply 2 by 1 and place the product in place of the first wildcard in the second row:
2 1 1 0 2 # −7 # # −6 # #
We add the numbers in the third column.” We put that number in the top left slot:
2 1 # 0 # # −7 # # −6 # #
Next.424 Cubic Equations in Real Numbers
To set up the synthetic division “grid. We put those numbers into the top row from left to right.

Here’s the sequence of steps:
4 1 # 0 # # −7 # # −6 # #
. we fail. in the bottom row:
2 1 1 0 2 2 −7 4 −3 −6 −6 −12
This final number is called the remainder. Then the process goes along with the new numbers. writing the sum in the bottom row:
2 1 1 0 2 2 −7 4 −3 −6 # #
Do you see the pattern? Next. Let’s write this last number. and write the product under the entry −6:
2 1 1 0 2 2 −7 4 −3 −6 −6 #
Adding the numbers in the last column yields the final result of our test.Polynomial Equation of Third Degree
425
We multiply this sum by the “test root” and place the product beneath the entry −7:
2 1 1 0 2 2 −7 4 # −6 # #
We add the numbers in the fourth column. We put a bold numeral 4 in the top left slot.
We try. in the same way as before. We must try another “test root” and go through this ritual again. We want the remainder to be 0! That is always the ultimate goal of synthetic division when we’re looking for a binomial factor of a cubic equation. the news is not good. What does this particular result mean? Well. we multiply the “test root” by −3. we learn Let’s try 4 as our “test root” this time.

. and to get a little complementary credit. can be found by solving the quadratic equation x 2 + 3x + 2 = 0 The roots of this equation turn out to be x = −1 and x = −2. The first three of these.” x = −3/2.428 Cubic Equations in Real Numbers
What are the other roots? We can write the binomial-trinomial form of a cubic equation straightaway. We can therefore write the binomial-trinomial version of the original cubic as
(x − 3)(x 2 + 3x + 2) = 0 To be certain of this. in the above example. show that −3/2 is the only real root. Take a close look at the numbers in the bottom line of the last step in the synthetic division process for the “test root” of 3.−1. That gives us three real roots for the original cubic: x = 3 or x = −1 or x = −2 and a real-number solution set X = {3. from the results of that process. Those numbers are 1. The factor theorem tells us so. if they exist. and 0. all irrational numbers? What if all the methods to “crack a cubic” that we’ve seen in this chapter fail us? There are other schemes available. that (x − 3) is the binomial factor.−2}. but in some cases it does not. 3. What if the real roots of a cubic are all complicated fractions—or worse. once we have performed synthetic division on the original cubic and managed to come up with a remainder of 0. some of which will be covered in the next chapter. in the order shown. We know. 2. you can multiply the left side of the equation out and see that it produces the original polynomial. For some more complementary credit.
Here’s a challenge!
Consider the following cubic in polynomial standard form: 6x 3 + 13x 2 + 8x + 3 Show by synthetic division that −3/2 is a real root of this equation. So the trinomial factor is (x 2 + 3x + 2). let’s set up the synthetic division array with the “test root.
Are you confused?
Synthetic division sometimes works out nicely. The other two real roots. You can verify this fact as another complementary-credit exercise. Then. you can substitute each of these roots for x in the original cubic to demonstrate that they are correct.
Solution
First. and the coefficients in the top row. are the coefficients and the stand-alone constant in the trinomial factor of the cubic equation.

430 Cubic Equations in Real Numbers
That’s (x + 3/2). the trinomial factor is 0. the coefficient of x 2 is 6. The only way the original cubic polynomial can attain the value 0 is if either the binomial factor is 0. as follows: d = b22 − 4a2c = 42 − 4 × 6 × 2 = 16 − 48 = −32 If we set the trinomial factor equal to 0 to get a quadratic equation. and the standalone constant is 2. In the chapter text. Multiply out the following equation to obtain a polynomial cubic: (ax + b)3 = 0 where x is the variable. 4. the coefficient of x is 4. by all means try it! 1. we can find the discriminant. 2.
. and a and b are real numbers with a ≠ 0. If we let a2 = 6. Here’s the binomial-trinomial cubic: (x + 3/2)(6x 2 + 4x + 2) = 0 Let’s examine the trinomial factor. then that quadratic has no real roots because d < 0. b2 = 4. In the trinomial factor. It follows that x = −3/2 is the only real root of the original cubic equation. and 2 appear. Multiply out the following equation to get a polynomial cubic: (31/2x − 121/2)3 = 0 3. The binomial becomes 0 if and only if x = −3/2. d. or both. in the bottom row before the remainder 0.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems. C. we solved this cubic in binomial factor form: (3x + 2)(5x + 6)(−7x − 1) = 0 We found that the real roots are x = −2/3 or x = −6/5 or x = −1/7 Multiply this equation out to get it into the polynomial standard form. If you think you can solve a particular problem in a quicker or better way than you see there. Don’t hurry! You’ll find worked-out answers in App. We’ve just discovered that no real number x can make the trinomial become 0. We know this because 6. and c = 2. The solutions in the appendix may not represent the only way a problem can be figured out. in that order.

3. One of the “challenges” in the text required that we find the roots of the following cubic equation in binomial-trinomial form: (3x + 5)(16x 2 − 56x + 49) = 0 We found that the real roots are x = −5/3 or x = 7/4 Substitute these roots into the original equation. verify that x = 5 is a real root of this equation. and go through the arithmetic to verify that they’re accurate. 9. 3. The final “challenge” in the text revealed that x = −3/2 is the only real root of the following cubic as expressed in the binomial-trinomial form: (x + 3/2)(6x2 + 4x + 2) = 0 How many real roots will the cubic have if the coefficient of x in the trinomial is changed from 4 to −4? What will the new real roots be.Practice Exercises
431
4. in place of x in the polynomial equation derived in the solution to Prob. Then state the solution set. if any exist. Verify that these three roots work in that equation. 10. Multiply out the general binomial-trinomial equation to get an equation in polynomial standard form for a cubic: (a1x + b1)(a2x 2 + b2x + c) = 0 6. one by one. Substitute the real roots that we found for the equation stated in Prob. By means of synthetic division. 8. if there are any?
. Consider the following cubic equation in polynomial standard form: −9x 3 + 21x 2 + 104x + 80 = 0 8. find the other root or roots of the original cubic. How can we tell from the coefficients and the constants alone how many real roots the equation stated in Prob. 5. Using the quadratic formula on the results of the synthetic division performed in the solution to Prob. 5 has (before multiplying it out)? 7.

a formidable polynomial equation can be reduced to binomial to the nth form. 7. we can find the real root by considering the binomial as a first-degree equation. is useful. which are of degree 4. if you can carry it out. and 17. Then it’s easy to solve! What’s the real root? Whenever we see an equation in binomial to the nth form.” Let’s look at three equations that have identical solution sets. we get
2x − 3 = 0 6x + 1 = 0 23x + 77 = 0 −7x − 12 = 0 −118x + 59 = 0 −3x = 0 These all resolve easily. Once in awhile. but the result usually looks more complicated. x = 3/2. The first-degree equation 2x − 3 = 0 has one real solution.Binomial to the nth Power
433
(−7x − 12)7 = 0 (−118x + 59)13 = 0 (−3x)17 = 0 In the last case. 13. and 17 respectively. 13. the stand-alone constant is 0. as would happen in the fifth example above. “What’s all this fuss about root multiplicity? If an equation has one real root. In some instances. The fourth-degree equation (2x − 3)4 = 0 has a single real root. But there’s something about the fourth-degree equation that makes it conceptually different than the first-degree equation. 5. 5. of multiplicity 4. 7. 6. But the reverse process. have one real root apiece. The original equations.
Multiplying out You can multiply out an equation in binomial to the nth form to get a polynomial equation. 6. “Not exactly.
Are you confused?
Perhaps you still wonder. When we do that with the above higherdegree equations. the multiplied-out equation becomes messy indeed. isn’t that all there is to be said about it?” The answer is. We can rewrite the fourth-degree equation as (2x − 3)(2x − 3)(2x − 3)(2x − 3) = 0
. x = 3/2.

434 Polynomial Equations in Real Numbers
If we substitute 3/2 for x here. “four times over. (x 2 + 2x + 1)2 = 0
Solution
Look closely at the trinomial. the single real root x = 3/2 exists “345 times over. we get (2 × 3/2 − 3)(2 × 3/2 − 3)(2 × 3/2 − 3)(2 × 3/2 − 3) = 0 which reduces to (3 − 3)(3 − 3)(3 − 3)(3 − 3) = 0 and further to 0×0×0×0=0 To carry out the substitution and simplification process completely. and the 345th-degree equations in this example. which is in the form of a trinomial squared. so it becomes the quadratic x 2 + 2x + 1 = 0 We can factor this to get (x + 1)2 = 0 If we substitute (x + 1)2 for the trinomial in the original equation. and state the multiplicity of each. 9 tells us that this is the same as (x + 1)(2×2) = 0 which can be simplified to (x + 1)4 = 0
.” The solution sets are identical for the first-degree. the fourth-degree. Suppose we set it equal to 0.” Now look at this: (2x − 3)345 = 0 Here. in effect. But the equations themselves are vastly different!
Here’s a challenge!
Consider the following equation. Find all the real roots. The root x = 3/2 exists. we must repeat it for each of the four binomials. we have [(x + 1)2]2 = 0 The product of powers rule from Chap.

.
Multiplying out You can multiply out an equation that appears in the binomial factor form to get a polynomial equation.Binomial Factors
435
We now have a binomial to the nth equation with n = 4. You can find each root by setting every binomial equal to 0 and then creating first-degree equations from them. We can find the solitary real root by setting the binomial equal to 0: x+1=0 This first-degree equation resolves to x = −1. b3. then its root has multiplicity equal to that power.
Binomial Factors
A higher-degree equation can appear as a product of binomials. go ahead! You’ll end up with equations that are difficult to solve for anyone who comes across them for the first time. b2. two of the stand-alone constants are equal to 0. The single root has multiplicity 4.. a2. . . That’s the equivalent of repeating those binomials in the products by the number of times the power indicates. In the last equation.. a3. If a binomial is raised to a power. Here are some examples of equations in this form where n > 3: (x + 1)(x − 2)(x + 3)(x − 4) = 0 (x + 1)(x − 2)2(x + 3)3(x − 4)4 = 0 (3x + 7)6(4x − 5)2 = 0 (−x + 1)(−7x + 2)7 = 0 (−3x + 21)2(−8x + 5)3 = 0 (x + 6)(2x − 5)2(7x)(−3x) = 0 In all but the first of these equations. Consider the equation
(a1x + b1)(a2x + b2)(a3x + b3) ··· (anx + bn) = 0 This is the binomial factor form for an equation of degree n. some of the binomials are raised to powers. bn are real stand-alone constants. If you want to do this for each of the six equations in the previous paragraph. and b1. Such an equation has one real root for each factor. Suppose that a1. an are nonzero real coefficients of x. The real solution set of the original trinomial-squared equation is X = {−1}. Each root has multiplicity equal to the number of times its binomial appears in the product.
The general form Let x be a real-number variable...

436 Polynomial Equations in Real Numbers
What are the real roots? As we search for real roots to a higher-degree equation in binomial factor form. set it equal to 0. one by one.
Are you confused?
Does the concept of multiplicity still seem esoteric? In the example we just finished. and the root x = 4 has multiplicity 4. 4}. and then solve the resulting first-degree equations: x+1=0 x−2=0 x+3=0 x−4=0 Any value of x that satisfies one of these is a root of the higher-degree equation. That means the real roots of the higher-degree equation are the same. That means we can set the binomial factors equal to 0. Grouped according to their constants. those factors are (x + 1) (x − 2)(x − 2) (x + 3)(x + 3)(x + 3) (x − 4)(x − 4)(x − 4)(x − 4)
. we get the same four first-degree equations as before. But three of the four roots have multiplicity greater than 1. −3. There are four such values: x = −1 or x = 2 or x = −3 or x = 4 so the real solution set of the higher-degree equation is X = {−1. it can help if we write out every binomial factor individually so none is raised to a power (other than the first power). Consider the first of the six binomial factor equations listed above. because three of the four binomials have exponents attached: (x + 1)(x − 2)2(x + 3)3(x − 4)4 = 0 If we remove the exponents from the binomials and set each equal to 0. Now consider another higher-degree equation. the root x = −3 has multiplicity 3. and then solve each of them to get all the real roots. The root x = 2 has multiplicity 2. too. 2. This one is a little tricky. Let’s find all the real roots of
(x + 1)(x − 2)(x + 3)(x − 4) = 0 We take each binomial individually. we know we’ve “made a hit” when any of the factors becomes 0.

twice when x = 2. The root −6 has multiplicity 1. These facts can be clarified by stating the original equation as (x + 6)(2x − 5)(2x − 5)(7x)(−3x) = 0 The degree of the original equation is the sum of the exponents attached to the factors. In this example. That’s 1 + 2 + 3 + 4 = 10. What is the degree of the equation? (x + 6)(2x − 5)2(7x)(−3x) = 0
Solution
We take each binomial individually. three times when x = −3. −3. whether we write the equation in the original form or in the fully expanded binomial factor form. On that basis. 0}.
.Binomial Factors
437
We have a root in every single case where one of these factors becomes equal to 0. the degree of the original equation is equal to the sum of the exponents attached to the binomial factors.
Here’s a challenge!
State the real roots of the following equation. The root 5/2 has multiplicity 2. 2. There are not four such instances here. 4} Incidentally. This is true even though the real solution set has only four elements: X = {−1. 5/2. Also state the real solution set X and the multiplicity of each root. set it equal to 0. indicating that it’s a fifth-degree equation. and four times when x = 4. that sum is 5. but 10! We “hit the target” once when x = −1. and then solve the resulting first-degree equations: x+6=0 2x − 5 = 0 7x = 0 −3x = 0 The real roots are x = −6 or x = 5/2 or x = 0 and the real solution set is X = {−6. we can immediately see that (x + 1)(x − 2)2(x + 3)3(x − 4)4 = 0 is a 10th-degree equation in the variable x. The root 0 has multiplicity 2.

If n = 1. The coefficient an.. Here are some examples:
x 2 = 2x + 3x 7 − 4x 15 x = 4x 21 − 7x 17 + 2x 11 + 2x 7 8 x + 4x 6 + 7x 4 − x 2 + 3 = x + x 3 + x 5 + x 7 The only requirements for membership in the “single-variable nth-degree equation club” are that the equation be convertible into polynomial standard form. we have a higher-degree equation. if n > 3. and that n be a positive integer.. . an are coefficients. we have a first-degree equation. a2.
General polynomial equation The polynomial standard form of a higher-degree equation can be written as
an x n + an-1x n-1 + an-2x n-2 + ··· + a1x + b = 0 where a1. if n = 2. The right side of the equals sign has 0 all by itself.
. can never be 0 in an nth-degree polynomial equation. If you set an = 0. some of the coefficients are equal to 0. a3. b is the stand-alone constant. we have a cubic. by which x n is multiplied. many single-variable equations can be morphed into the polynomial standard form.438 Polynomial Equations in Real Numbers
Polynomial Standard Form
A higher-degree equation in polynomial standard form contains a sum of multiples of the variable raised to powers in descending order on the left side of the equals sign. we have a quadratic. Here are some examples: 6x 4 − 3x 3 + 3x 2 + 2x + 5 = 0 3x 5 − 4x 3 = 0 −7x 7 − 5x 4 + 3x 3 − x 2 − 29 = 0 −4x 11 = 0 In all but the first of these equations. you end up with 0x n + an-1x n-1 + an-2x n-2 + ··· + a1x + b = 0 That’s the polynomial standard form for a single-variable equation of degree n − 1: an-1x n-1 + an-2x n-2 + ··· + a1x + b = 0
Mutants in the n th degree As you can imagine. and n is a positive integer greater than 3. if n = 3.

In part. you might wonder how long you should keep trying before you give up (or let your computer take over). and if we can discover a lower bound that’s small enough.Digging for Real Roots
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Digging for Real Roots
Now that we’ve learned how to recognize a polynomial equation. You’ve resolved the mystery. In this context. We can identify an interval that contains all the elements in the real solution set if we can discover an upper bound that’s big enough. We’ve been told to find the real roots. Anyone but the purest mathematician would likely concede that these types of situations lend themselves to computer programming. we consider only non-inclusive bounds. The next few sections offer some ways to look for the roots of higher-degree equations.
Bounds for real roots The real roots of a polynomial equation always lie between two extremes. it’s time to think about finding the real roots of such an equation. let’s look for the real roots of
x 4 − 2x 3 − 13x 2 + 14x + 24 = 0 This equation was built up from factors. it depends how much time and patience you have. The roots are x = −1 or x = 2 or x = −3 or x = 4 Now imagine that we’re looking at the polynomial version of this equation.
A prefabricated problem To illustrate how the real roots can be sought when we’re confronted with a polynomial equation. but all of the complex roots. Here’s what it looks like in binomial factor form: (x + 1)(x − 2)(x + 3)(x − 4) = 0 We solved this awhile ago. That includes not only the real roots. That means an upper bound must be greater than the largest real root of the equation. you can terminate your quest.
. But there are no guarantees. Here’s an important principle to keep in mind:
• A polynomial equation can never have more roots than its degree. There is nothing more to find. If you’re working on a polynomial equation of degree n and you’ve found n roots.
How many roots? When you embark on a quest to find the real roots of a polynomial equation. if any exist. and a lower bound must be less than the smallest real root of the equation. and we’ve never seen it in any other form.

that’s 24.” we can tell if it’s an upper bound for the real solution set by looking at the values we get in the last row. 14. and 24 in order of descending powers of x. In this case. −13. Let’s set up a synthetic division array for this equation:
# 1 # −2 # # −13 # # 14 # # 24 # #
If we plug in a positive real number as a “test root. It seems reasonable that this might be larger than or equal to all the real roots. If we get a nonzero remainder and none of the numbers in the last row are negative. then our “test root” is an upper bound.440 Polynomial Equations in Real Numbers
Finding an upper bound How can we find an upper bound for the roots of the polynomial equation under investigation? Here’s the equation again:
x 4 − 2x 3 − 13x 2 + 14x + 24 = 0 The coefficients and constant are 1. Let’s take absolute values of the coefficients and constant in the equation and pick out the largest. −2. Let’s input 24 and see what happens:
24 1 # 24 1 1 24 1 1 24 1 1 24 1 1 −2 # # −2 # # −2 24 # −2 24 22 −2 24 22 −13 # # −13 # # −13 # # −13 # # −13 528 # 14 # # 14 # # 14 # # 14 # # 14 # # 24 # # 24 # # 24 # # 24 # # 24 # #
.

The fact that the numbers increase so fast (we might even say that they “blow up”) suggests that 24 is a much larger than the largest root of the equation. get a nonzero remainder. Let’s take the absolute values of the coefficients and constant. That’s −24. and discover that the numbers in the last row alternate between positive and negative. Let’s input it to the array and find out:
−24 1 # −2 # # −13 # # 14 # # 24 # #
.360 12. Therefore. but still positive. 24 is an upper bound for the real solution set. it tells us that we’ve found a lower bound for the real solution set. we know that we’re inputting upper bounds. We can try something smaller.
Finding a lower bound If we plug in a negative number as a “test root. and do the synthetic division again.976 #
441
24
1 1
−2 24 22
−13 528 515
14 12.360 # 14 12.374
24 296.Digging for Real Roots 24 1 1 24 1 1 24 1 1 24 1 1 −2 24 22 −2 24 22 −2 24 22 −2 24 22 −13 528 515 −13 528 515 −13 528 515 −13 528 515 14 # # 14 12.374 14 12. It’s a good bet that this is smaller than all the real roots.” grind out the synthetic division process. and pick out the negative of the largest result. As long as we input positive “test roots” and never see a negative number in the last row.360 12.360 12.374 24 # # 24 # # 24 # # 24 296.000
None of the numbers in the bottom row are negative.976 297.

but still negative.. In this particular example. Suppose we gradually reduce the positive “test root” and gradually increase the negative “test root” in the synthetic division array
# 1 # −2 # # −13 # # 14 # # 24 # #
If we use integers for simplicity. an. . we know that we are in the interval containing the real roots. multiply the equation through by the smallest constant that will turn all the numbers a1.. or a negative “test root” that fails to cause the numbers in the bottom line to alternate in sign. b is a nonzero rational constant. a3. We can try something larger. and n is a positive integer greater than 3.
. The fact that the absolute values diverge rapidly from 0 suggests that our input number is far smaller than necessary. If that is not the case. a2. Call these by the general name m. • Make certain that all the numbers a1. a3. an are nonzero rational-number coefficients of the variable x. . .Digging for Real Roots
443
The numbers in the last row alternate in sign. Call these by the general name n.. If b ≠ 0. telling us that −24 is a lower bound for the real solution set. then you can go through the following steps. a2. an. Call them by the general name r.. we’ll eventually get down to a smallest upper bound. we know that we’re inputting lower bounds. but you will at least know that 0 is a root. and b into integers.
Narrowing the interval We can reduce the size of the interval containing the real solutions by repeatedly testing positive numbers as upper bounds. a3. • Find all the positive and negative integer factors of an.
Rational roots There’s a lengthy but straightforward process you can use to find all the rational roots of a polynomial equation in the standard form
an x n + an-1x n-1 + an-2x n-2 + · · + a1x + b = 0 where a1. then you can’t use the process. and we’ll also get up to a greatest lower bound. we can test numbers between those bounds to look for real roots. and by repeatedly testing negative numbers as lower bounds. • Write down all the possible ratios m /n. we’ll get a smallest upper bound of 5 and a greatest lower bound of −4. the stand-alone constant. the coefficient of xn (sometimes called the leading coefficient).. a2. As long as we plug in negative “test roots” and get numbers in the last line that alternate in sign. Once we get a positive “test root” that produces a negative number anywhere in the bottom line. and do the synthetic division again. At that point. • Find all the positive and negative integer factors of b.. and b are integers. If b = 0.

• The positive and negative integer factors of the stand-alone constant. • If you’re lucky. What can we do to find those roots?” The best answer is. you’ll know that none of the roots of that equation are rational. and so on. are all the integers m that divide 24 without remainders. and for which we know the smallest upper bound is 5 and the greatest lower bound is −4. r2.444 Polynomial Equations in Real Numbers
• With synthetic division. (x − r2). If you set that factor equal to 0 to form a polynomial equation.
Here’s a challenge!
Use the above-described procedure to find all the rational roots of the polynomial equation we contrived earlier in this chapter. then the original polynomial equation has no rational roots. 2. 12. • All the coefficients. • If you’re unlucky. that equation is x 4 − 2x 3 − 13x 2 + 14x + 24 = 0
Solution
Here is an outline of the process. and then use the computer to approximate the zeros of that function. are integers. They might even be complex.
. to verify. you’ll end up with one or more binomial factors and a quadratic factor. • List all of the rational roots found after carrying out the preceding steps. you’ll be stuck with one or more binomial factors and a cubic or higher-degree factor. Call them r1. you must wonder. you’ll get a remainder of 0 at the end of the synthetic division process. particularly the synthetic division problems. check every r to see if it is a root of the original polynomial equation. You might want to work out the arithmetic. If you find a root. see how many times that graph crosses the x axis. you’ll end up with an equation in binomial to the nth form. Some might even be complex. but you should not expect the task to be easy. “We can use a computer program to generate an approximate graph of the function produced by the polynomial equation. 4.
Are you confused?
At this point. then every one of those numbers is a rational root of the equation. Once again. • Create binomials of the form (x − r1).” This method will not allow us to find non-real roots. 24. That factor can be set equal to 0. (x − r3). • If one or more of the ratios r produces a remainder of 0. and then the quadratic formula can be used to find its roots. and 1. so we don’t have to multiply the equation through by anything. 6. or an equation in binomial factor form. as well as the stand-alone constant. • If you’re less lucky. These numbers are 24. and so on. along with all their negatives. and its associated equation has some irrational roots. “Suppose we’re left with a cubic or higher-degree polynomial as one of the factors. Neither of those roots will be rational. Each of these binomials is a factor of the original equation. r3. • If none of the numbers r produces a remainder of 0. You can try to solve it. 3. 8.

3? What is the multiplicity in each case? 5. • We get a remainder of 0 when r = 4. −2. 2. 4. That leaves us with rational numbers r of 4. Don’t hurry! You’ll find worked-out answers in App. What is the degree of the equation? (x + 4)(2x − 8)2(x /3 + 12)3 = 0
. Now we know that every one of those numbers is a rational root of the equation. 2. That includes not only the rational roots. along with all their negatives. 12. and −3 to synthetic division arrays. Imagine that we’ve narrowed down the interval by doing synthetic division repeatedly. • Now let’s cheat a little. Also state the real solution set X and the multiplicity of each root. What are the real roots for each of the equations stated in Prob. • All the possible ratios m /n are the same as the integers m: 24. There are no more rational roots to
find! In fact. If you think you can solve a particular problem in a quicker or better way than you see there. Remember: A polynomial equation can never have more roots than its degree.
Practice Exercises
This is an open-book quiz.Practice Exercises
445
• There are only two integer factors n of the leading coefficient: 1 and −1. You may (and should) refer to the text as you solve these problems. by all means try it! 1. C. 2. or r = −3. (a) (x 2 − 3x + 2)2 = 0 (b) (−3x 2 − 5x + 2)5 = 0 (c) (4x 2 − 9)3 = 0 4. State the real roots of the following equation. What are the real roots for each of the equations stated in Prob. Rewrite each of the following equations in binomial to the nth form. • We input 4. −1. 1? What is the multiplicity in each case? 3. but the irrational and complex roots. −1. r = 2. What is the degree of the equation? (x − 3/2)2(2x − 7)2(7x)3(−3x + 5)5 = 0 6. Also state the real solution set X and the multiplicity of each root. and −3 to check as possible roots. and see if we get a remainder of 0 for any of them. 1. 6. 1. Rewrite each of the following equations in binomial factor form. these are all the roots of any sort. r = −1. 3. • We have found four rational roots for a fourth-degree equation. finding a smallest upper bound of 5 and a greatest lower bound of −4. The solutions in the appendix may not represent the only way a problem can be figured out. 3. State the real roots of the following equation. and 1. 3. −2. (a) (x 2 + 6x + 9)2 = 0 (b) (x 2 − 4x + 4)3 = 0 (c) (16x 2 − 24x + 9)4 = 0 2. 8.

9. 9.446 Polynomial Equations in Real Numbers
7. Determine all the rational roots of the equation stated in Prob. Using synthetic division. Determine all the rational roots of the equation stated in Prob. 7. find a lower bound and an upper bound for the interval containing all the real roots of the equation 3x 5 − 3x 2 + 2x − 2 = 0 10.
. Using synthetic division. find a lower bound and an upper bound for the interval containing all the real roots of the equation 2x 5 − 3x 3 − 2x + 2 = 0 8.

so we don’t have to manipulate it. we mix When we mix the independent-variable parts of the above functions. 22 and 23. The simpler of the two is y = 2x + 1 For x = −2. because this equation can be factored into
(x + 2)(x − 3) = 0 The roots are found by solving the two first-degree equations x+2=0 and x−3=0 giving us x = −2 or x = 3. we have y = 2 × (−2) + 1 = −4 + 1 = −3
. getting y = 2x + 1 We now have two functions in which x is the independent variable and y is the dependent variable.
Next. The linear equation can be rearranged so it appears as a function of x by adding y to each side and then transposing the left and right sides. we solve We’ve derived a quadratic equation that can be solved using any of the methods from Chaps. then the quadratic equation is already expressed as a function of x. we have
x 2 + x − 5 = 2x + 1 We can subtract 1 from both sides to obtain x 2 + x − 6 = 2x Then we can subtract 2x from both sides to get x2 − x − 6 = 0
Next.448 More Two-by-Two Systemss
If we call x the independent variable. We’re lucky here. We can substitute these two values for x into either of the morphed original equations to obtain corresponding values for y.

Change the subtractions to negative additions. but we must use letter constants instead of specific numbers. we’ll plot these graphs. You’ll see that they intersect at the points corresponding to the ordered pairs (−2.450 More Two-by-Two Systemss
Are you confused?
We’ve found two ordered pairs (x.
Solution
We can follow the same procedure as we did for the example we just solved. distributive. followed by left-to-right transposition. a2. These two equations are ready to mix. allow us to rearrange this equation to get a2x 2 + (b2 − a1)x + (c − b1) = 0 If you’re confused by this rearrangement. b2. Consider these two functions of x : y = a1x + b1 and y = a2x 2 + b2x + c Derive a general formula for solving this linear-quadratic system. move things around. b1. In the next chapter. and then change them back again. and grouping principles. y) that solve the above pair of equations as a two-by-two system. then check it out for yourself.
Here’s a challenge!
Suppose that a1. obtaining 0 = a2x 2 + b2x + c − b1 − a1x The commutative. How do we know that these are the only two solutions for this system? We can demonstrate this visually for the real-number solutions by graphing both equations together on the coordinate plane. The linear equation shows up as a straight line. we get a1x + b1 = a2x 2 + b2x + c We can subtract b1 from each side to get a1x = a2x 2 + b2x + c − b1 We can then subtract a1x from each side. When we set the right sides equal to each other. and neither a1 nor a2 are equal to 0. or if you aren’t convinced that it’s correct. −3) and (3.
. but nowhere else. 7). and c are real numbers. and the quadratic shows up as a parabola. and we can’t simplify the expressions as much. we have no morphing to do.

and then reversing the positions of a1 and b2 inside the parentheses. Let’s state the quadratic formula once again as we originally learned it. the linear function is less messy than the quadratic. We can simplify this slightly by getting rid of the minus sign in the first expression on the right side of the equals sign. y2). but will not make the formula easier to use. let’s make these substitutions in the classical version: • Write a2 in place of a • Write (b2 − a1) in place of b • Write (c − b1) in place of c When we do that. That means we can solve it directly with the quadratic formula. we can plug them into either of the original functions to obtain the y-values. we get x = {−(b2 − a1) ± [(b2 − a1)2 − 4a2(c − b1)]1/2} / (2a2) The roots defined by this formula give us the x-values for the solution of the original linear-quadratic system. In this case. y1) and (x2.
Two Quadratics
Now let’s solve a two-by-two system consisting of these quadratic equations in the two variables x and y : 4x 2 + 6x + 2y + 8 = 0
. then y1 = a1x1 + b1 and y2 = a1x2 + b1 The solutions of the whole system can be written as the ordered pairs (x1. For the general quadratic ax 2 + bx + c = 0 the roots are given by x = [−b ± (b2 − 4ac)1/2] / (2a) To make the quadratic formula work in the current equation. When we’ve found the x-values of the solutions. getting x = {(a1 − b2) ± [(b2 − a1)2 − 4a2(c − b1)]1/2} / (2a2) Any attempt to further simplify this will eliminate some grouping symbols.Two Quadratics
451
Now we have a quadratic equation in standard form. If we call the x-values x1 and x2.

getting −y = 3x 2 + 5x − 11 Multiplying through by −1. a multiple of y can be separated out and placed alone on the left side of the equals sign. we can subtract 2y from each side and then transpose the sides to get
−2y = 4x 2 + 6x + 8 Dividing through by −2. obtaining x 2 + 2x − 15 = 0 which is a quadratic equation in polynomial standard form. we get the function y = −2x 2 − 3x − 4 In the second original equation. In the first equation. we get y = −3x 2 − 5x + 11
Next. we get a single equation in one variable:
−2x 2 − 3x − 4 = −3x 2 − 5x + 11 We can add the quantity (3x 2 + 5x − 11) to each side. we can subtract y from each side and then transpose the sides.
Next. we mix When we directly mix the right sides of the above two quadratic functions. It does not take long to figure out that the above quadratic is equivalent to
(x + 5)(x − 3) = 0 The roots are found by solving the equations x+5=0
. we morph In both of these equations. we solve We now have an equation that can be easily factored. producing quadratic functions of x.452 More Two-by-Two Systemss
and 3x 2 + y + 5x − 11 = 0
First.

To minimize the risk of error. (Have you memorized it yet?) When we see the quadratic equation ax 2 + bx + c = 0 the roots are given by x = [−b ± (b2 − 4ac)1/2] / (2a) To solve the equation at hand. here it is. yet one more time. so we can mix the right sides directly without having to manipulate.Two Quadratics
455
Solution
Both of these equations are presented as functions of x . let’s do negative additions. we get x = {−(b1 − b2) ± [(b1 − b2)2 − 4(a1 − a2)(c1 − c2)]1/2} / [2(a1 − a2)] We can simplify this slightly by getting rid of the minus sign in the first expression on the right side of the equals sign. getting a1x 2 + b1x + c1 + (−a2x 2) + (−b2x) + (−c2) = 0 Using the commutative law for addition. we can rewrite this as a1x 2 + (−a2x 2) + b1x + (−b2x) + c1 + (−c2) = 0 The distributive law of multiplication over addition allows us to group the coefficients and the constants to get [a1 + (−a2)]x 2 + [b1 + (−b2)]x + [c1 + (−c2)] = 0 Changing the negative additions back to subtractions. and change the negative additions to subtractions after we’ve put everything in the proper order. we obtain (a1 − a2)x 2 + (b1 − b2)x + (c1 − c2) = 0 We can solve this equation with the quadratic formula. For reference. When we do that. and then reversing the positions of b1 and b2 inside the parentheses. in its classical form. we get a1x 2 + b1x + c1 = a2x 2 + b2x + c2 The next several steps are tricky. rearrange and regroup the symbols. We can add the negatives of a2x2. b2x. we can make these substitutions in the classical version of the quadratic formula: • Write (a1 − a2) in place of a • Write (b1 − b2) in place of b • Write (c1 − c2) in place of c When we make these changes in “copy-and-paste” fashion. and c2 to each side. We can also multiply out
. It’s easy to make mistakes with the signs and the grouping.

But there’s no good reason to expand the products of the binomials. Let’s express the formula as x = {(b2 − b1) ± [(b1 − b2)2 − 4(a1 − a2)(c1 − c2)]1/2} / (2a1 − 2a2) After we’ve found the x-values of the solutions by plugging in the coefficients.
Enter the Cubic
Let’s solve a two-by-two system in which one equation is linear and the other is cubic. we can add 7 to each side.456 More Two-by-Two Systems
the denominator. obtaining y as a function of x : y = x 3 + 6x 2 + 14x + 7 In the second equation. and going through the arithmetic. That would produce a formula with fewer grouping symbols. we can divide through by −2 to obtain 3x − y = −1 Adding 1 to each side gives us 3x − y + 1 = 0
. If we call the x-values x1 and x2. getting
x 3 + 6x 2 + 14x − y + 7 = 0 Then we can add y to each side and transpose the equation left-to-right. we morph Again. it appears as if we ought to let x be the independent variable. y1) and (x2. y2). Consider these: x 3 + 6x 2 + 14x − y = −7 and −6x + 2y = 2
First. plugging in the constants. but the arithmetic would be more cumbersome in practical use. we can put those x-values into either of the original two functions and calculate the y-values. In the first equation. and then derive two functions of that variable. and if we use the first of the original functions. we have y1 = a1x12+ b1x1 + c1 and y2 = a1x22 + b1x2 + c1 The solutions of the whole system can be written as ordered pairs in the form (x1.

we have y = 3x + 1 = 3 × (−2) + 1 = −6 + 1 = −5
. y) = (−1. we get x 3 + 6x 2 + 11x + 6 = 0 This is a straightforward cubic equation in polynomial standard form. The y-values can be found by plugging these roots into either of the original functions.
Next. but we can use the techniques from Chap. Those roots are x = −1. Ultimately. 25 to solve it. it’s the less messy of the two! For x = −1. we mix When we mix the independent-variable parts of the above functions. When x = −2. We can use synthetic division several times to obtain factors. −2). we find that the cubic factors into
(x + 1)(x + 2)(x + 3) = 0 The roots can be found by solving the three equations we get when we set each binomial equal to 0. and x = −3. we obtain one equation in one variable:
x 3 + 6x 2 + 14x + 7 = 3x + 1 If we subtract the quantity (3x + 1) from both sides. getting the function y = 3x + 1
Next. Let’s use the linear one. we have y = 3x + 1 = 3 × (−1) + 1 = −3 + 1 = −2 Now we know that our first solution is (x. The roots aren’t obvious from casual inspection. x = −2.Enter the Cubic
457
We can add y to each side and transpose the equation left-to-right. our mission is to find its roots. we solve Now that we have derived a cubic equation in one variable.

or both. You’ll see this happen as you work out the last two practice exercises at the end of this chapter. it does! If the equation we create by mixing has a root with multiplicity of 2 or more. so we have no morphing to do. we check (−3. a lot of trial and error. We can directly mix the right sides to get 5x 3 + 3x 2 + 5x + 7 = 2x 3 + x 2 + 2x + 5 Let’s subtract the entire right side of this equation. from both sides. We are lucky in one respect. we check (−2.
Here’s a challenge!
Solve these cubic equations as a two-by-two system: y = 5x 3 + 3x 2 + 5x + 7 and y = 2x 3 + x 2 + 2x + 5
Solution
Solving this problem requires some keen intuition. giving us 3x 3 + 2x 2 + 3x + 2 = 0
.Enter the Cubic
459
Next. Do you wonder if the same concept applies to the solutions of two-by-two systems when at least one of the equations is of degree 2 or more? Well. −8) in the second original equation: −6x + 2y = 2 −6 × (−3) + 2 × (−8) = 2 18 + (−16) = 2 2=2
Are you confused (or bemused)?
Think back to the notion of multiplicity for the roots of certain quadratic. at least: These equations are already functions of x. the corresponding solution of the whole system has the same multiplicity. −5) in the second original equation: −6x + 2y = 2 −6 × (−2) + 2 × (−5) = 2 12 + (−10) = 2 2=2 Finishing up. cubic. That changes each term on the left and sets the right side equal to 0. as a single quantity. and higher-degree equations.

and then define y as a function of x in each equation: 3x + y − 1 = 0 and 2x 2 − y + 1 = 0 2. If you think you can solve a particular problem in a quicker or better way than you see there. That’s all right for now. You may (and should) refer to the text as you solve these problems.
461
And finally . come back and check them for complementary credit. including the complexnumber solutions. Don’t hurry! You’ll find worked-out answers in App. including the complexnumber solutions. Check the solution(s) to Prob. when your mind is rested. 4). The solutions in the appendix may not represent the only way a problem can be figured out. 3. Tomorrow. and you’d prefer to take these solutions on faith. y) = (−j. Let x be the independent variable. Solve the following pair of equations as a two-by-two system. by all means try it! 1.Practice Exercises
= j 2 + (−1) + (−j 2) + 5 = j 2 + −j 2 + (−1) + 5 =0+4 =4 Our third solution is (x . Now it’s time to check these solutions to be sure that they work. Check the solution(s) to Prob. 3 in the original equations for correctness.. C. I can’t blame you if you’re weary of doing arithmetic.
. and then define y as a function of x in each equation: 3x + y − 1 = 0 and 2x 2 − 3x − y + 3 = 0 4. Solve the following pair of equations as a two-by-two system. if any.. 1 in the original equations for correctness. Let x be the independent variable. if any.
Practice Exercises
This is an open-book quiz.

9. Check the solution(s) to Prob. Solve the following pair of equations as a two-by-two system. and then define y as a function of x in each equation: x 2 + x − y = −1 and x 2 − 2x − y = 2 6. if any. if any. and then define y as a function of x in each equation: x2 + y = 0 and 2x 3 − y = 0 Explain why one of the solutions has multiplicity 2. and then define y as a function of x in each equation: 4x 3 + 2x 2 + 2x − 2y − 8 = 0 and 3x 3 − 2x 2 + 4x − y − 5 = 0 What is the multiplicity of each solution? 10. including the complexnumber solutions. if any. Let x be the independent variable. Let x be the independent variable. Solve the following pair of equations as a two-by-two system.
. 9 in the original equations for correctness. including the complexnumber solutions. Solve the following pair of equations as a two-by-two system. 7 in the original equations for correctness. 7. Let x be the independent variable. 5 in the original equations for correctness. Check the solution(s) to Prob. including the complexnumber solutions. Check the solution(s) to Prob.462 More Two-by-Two Systems
5. 8.

and a span of −7 to 25 for the dependent variable. We want to find ordered pairs that lie in the vicinity of the solutions.
Selected values for graphing the functions y = 2x + 1 and y = x 2 + x − 5. and somewhat larger than 3. and some values of the second function. we could use a strict Cartesian plane with each division on both axes equal to 1 unit. That means we should choose values of x that are somewhat less than −2. The right-hand column contains values of the quadratic function that we get for the indicated values of x. The middle column contains values of the linear function that we get when we input the chosen values of x. we can expect to get a good view with the solutions near the middle. 7) These solutions are written down in boldface. when we plot the graphs. But we must cover a span of −4 to 5 for the independent variable.
Next.
2x + 1 −7 −5 −3 1 3 7 9 11 x2 + x − 5 7 1 −3 −5 −3 7 15 25
x −4 −3 −2 0 1 3 4 5
Table 28-1 compares some values of x. The other points are chosen strategically. somewhere between −2 and 3. Then. −3) and (x. some values of the first function. we plot the solution(s) To plot the solutions of this system. the points corresponding to these solutions will be the points where the graphs intersect. so it makes sense to include them: (x. When we graph the functions. The left-hand column contains selected values of x.464 More Two-by-Two Graphss
Table 28-1. We calculate the values of the functions in the middle and right-hand columns by plugging in the values of x and going through the arithmetic.
Are you confused?
Do you wonder why we chose the values in Table 28-1 as we did? The two solutions can be tabulated easily. A true Cartesian graph having that span would be as big as a road map!
. Bold entries indicate real solutions. y) = (3. based on the function values we have found in Table 28-1. y) = (−2.

so we can see that the line has a slope of 2 and a y-intercept of 1. fill in its line or curve with a pen. and the other points will be scattered fairly well over the rest of it. 28-1. say. and it’s easy to get them confused.Linear and Quadratic
465
We could shrink it by making each division represent. but not strict Cartesian coordinates. the best approach is to use rectangular coordinates. the solution points get closer together. and then erase all the penciled-in points when the ink is dry. Imagine moving the straight line downward in Fig. we plot the two solution points. we can locate and plot the rest of the points in the table. and each increment on the y axis represent 5 units. Suppose we change the linear function so the y-intercept is −10. Our goal is only to get a good fit for the ranges of the values we determined when we created Table 28-1. we can use pencil and draw the points lightly. 28-1 shows the final result.
Solution
If we change the linear function so its graph doesn’t intersect the parabola. they’d be too close together. As we do this. fill in its line or curve with a pen.
Finally. Let’s make each increment on the x axis represent 1 unit. 28-1. we must remember the increments we’ve chosen for each axis. and assuming the slope of the line stays the same. That will put the line completely below the parabola. In situations like this. the solutions will be near the center of the coordinate area. eventually merging. When we make the axis increments different in size. the system has a single real solution with multiplicity 2. but that’s not important here. we plot the rest Once we’ve plotted the solution points as dots on the coordinate grid. The linear function is expressed in slope-intercept form. To begin. so we can erase them later. They aren’t the same. 5 units on both axes. As we draw the points. then the resulting two-by-two system will have no real solutions. so we can see that the quadratic function has an absolute minimum of approximately −5. draw the points for the second graph with a pencil.
Here’s a challenge!
By examining Fig. and a clear picture of how the graphs intersect. At the moment the two points become one. It’s also important to keep track of which graph goes with which function! We can draw the points for one graph with a pencil. describe how the linear function y = 2x + 1 (shown by the solid line) can be modified to produce a system with no real solutions. assuming that the quadratic function (shown by the dashed curve) stays the same. Fig. Each division on the y axis is 5 units. If we’ve chosen the axis increments wisely. But then the solutions wouldn’t show up well. If we do this on paper. and will give us the two-by-two system y = 2x − 10 and y = x2 + x − 5
. we distort the slopes and contours of the lines and curves.

but they’ll both be non-real complex numbers. On the x axis.466 More Two-by-Two Graphs
y
(3.
Are you confused?
Does the slope of the line in Fig.7)
x
(–2. On the y axis. each increment is 1 unit. The parabola would also look different. each increment is 5 units. Real-number solutions appear as points where the line and the curve intersect. 28-1 seem smaller than 2? In a way. That distorts the slopes and contours of the graphs.
If we solve this system.–3)
Figure 28-1 Graphs of y = 2x + 1 and y = x 2 + x − 5. the second function is graphed as a dashed curve. If we had used a true Cartesian plane. 27. it would seem “five times sharper. we’ll get two results. it is! In the algebraic sense the slope is 2. The first
function is graphed as a solid line.”
Two Quadratics
In the second example in Chap. we solved this system of quadratic equations in two variables: 4x 2 + 6x + 2y + 8 = 0 and 3x 2 + y + 5x − 11 = 0
. the line would have the steepness we should expect for a slope of 2 when drawn. you can solve the “revised” system and see for yourself. The increments on the vertical axis are five times as large as the ones on the horizontal axis. expanding everything horizontally (or compressing everything vertically) by a factor of 5. but in the geometric sense it’s only 2/5. For extra credit.

−39) and (x. −31) These solutions are written as bold numerals. 27.
Table 28-2. along with the results of plugging those values into the above functions and churning out the arithmetic. The other values are chosen to produce graph points in the vicinity of the solutions. We can choose a couple of x-values less than −5. and write them in the middle and right-hand columns of the table. two more between −5 and 3.Two Quadratics
467
Again. and then we manipulated the equations to obtain y as a function of x in both cases. we let x be the independent variable. which we determined in Chap. Selected values for graphing the functions y = −2x 2 − 3x − 4 and y = −3x 2 − 5x + 11. They are
(x. and two more larger than 3. That gave us y = −2x 2 − 3x − 4 and y = −3x 2 − 5x + 11
First. Bold entries indicate real solutions. Then we can calculate the values of the functions. we tabulate some points Table 28-2 shows some values of x. y) = (−5. We can start building the table by entering the two solutions.
x −10 −7 −5 −2 0 3 6 9 −2x 2 − 3x − 4 −174 −81 −39 −6 −4 −31 −94 −193 −3x 2 − 5x + 11 −239 −101 −39 9 11 −31 −127 −277
. y) = (3.

each increment is 2 units. We then plot the two solution points. In Fig. we plot the rest We can fill in the graphs by plotting the remaining points indicated in Table 28-2. Let’s make each increment on the x axis represent 2 units. we plot the solution(s) Once again. the second function is graphed as a dashed curve. fill in its curve with a pen.
. as shown in Fig. With six “hash marks” going out from 0 in each of the four directions along the axes.468 More Two-by-Two Graphs
y
(–5. let’s use rectangular coordinates. and the approximate curve for y = −3x 2 − 5x + 11 is a dashed parabola. On the x axis. wait for the ink to dry. it’s a good idea to draw the points for one graph with a pencil. Finally. Realnumber solutions appear as points where the curves intersect. fill in its curve with a pen. and finally run an eraser over the curves to get rid of the pencil marks.–31)
x
Figure 28-2
Graphs of y = −2x 2 − 3x − 4 and y = −3x 2 − 5x + 11. As before. 28-2. the span of absolute values for the input (the independent variable x) is from 0 to 10.–39)
(3.
Next. draw the points for the other graph with a pencil. the approximate graph for
y = −2x 2 − 3x − 4 is a solid parabola. each increment is 50 units. while the absolute values of the functions go as high as 277. that gives us absolute-value spans from 0 to 12 for x and 0 to 300 for y. 28-2. On the y axis. In this case. The first function is graphed as a solid curve. and each increment on the y axis represent 50 units.

Are you confused?
“How. Look again at the process we used in Chap. We can increase the value of the stand-alone constant. 28-2. “How much must we increase the constant in the first function to be sure that the solid parabola clears the dashed parabola?” We can find out using some creative algebra. 27 to derive an equation for the x-value of the solution to this system. We started with the two quadratic functions y = −2x 2 − 3x − 4 and y = −3x 2 − 5x + 11 When we mixed the right sides. indicating that the system has a single real solution with multiplicity 2. If we move the solid parabola upward beyond the “critical altitude.
Solution: Phase 1
We can move the solid parabola straight upward. the two intersection points get closer together.” the two parabolas no longer intersect.” the intersection points merge. we got −2x 2 − 3x − 4 = −3x 2 − 5x + 11 which simplified to x 2 + 2x − 15 = 0
. That increases the y-values of all the points without changing their x-values. assuming that the other quadratic function (shown by the dashed curve) stays the same. When the solid parabola reaches a certain “critical altitude. “can we change a quadratic function to move its parabola straight upward?” The answer is simple. and also assuming that the contour of the graph for the modified function stays the same.” you say. The solid parabola is not as “sharp” as the dashed one.Two Quadratics
469
Here’s a challenge!
By examining Fig.” you ask.
Solution: Phase 2
“All right. describe how the quadratic function y = −2x 2 − 3x − 4 (shown by the solid curve) can be modified to produce a system with no real solutions. As we do that. We can intuitively see this by comparing the “sharpness” of the two curves. Every point on the parabola is displaced straight upward by the same amount as every other point. leaving the rest of the equation unchanged. so the two curves diverge once we have raised the solid parabola high enough to completely clear the dashed one.

and the system has no real solutions. So let’s increase that constant by 17! When we do that. we get 0 = 22 − 4 × 1 × k which simplifies to 0 = 4 − 4k and further to 4k = 4 This resolves to k = 1. which is −15 right now.470 More Two-by-Two Graphs
Examine the discriminant d for this equation: d = 22 − 4 × 1 × (−15) = 4 − (−60) = 4 + 60 = 64 How can we change the stand-alone constant. the two solution points will merge in the graph. then d becomes negative. but displaced by 17 units upward in the coordinate plane. If use k as the letter constant. 28-2. In the quadratic we got by mixing. For extra credit. the function becomes y = −2x 2 − 3x + 13 The graph of this function is a parabola with the same contour as the original one. 28-2. and by substituting a letter constant for −15 in the above equation. let’s change the stand-alone constant k from −15 to 1.
Here’s another challenge!
By examining Fig. to bring the discriminant down to 0? We can find out by substituting 0 for d. That gives us x 2 + 2x + 1 = 0 If we increase the constant any more. Now we know that if we increase the stand-alone constant by 16 in the quadratic represented by the solid parabola in Fig. try solving the system y = −2x 2 − 3x + 13 and y = −3x 2 − 5x + 11 and see for yourself that it has no real solutions. describe how the quadratic function = −3x 2 − 5x + 11
.

and are written in bold. we tabulate some points Table 28-3 shows several different values of x. −5) (x. In this situation we got y = x 3 + 6x 2 + 14x + 7 and y = 3x + 1 We found these solutions: (x.Enter the Cubic
471
(shown by the dashed curve) can be modified to produce a system with no real solutions. Because the x-values for the solutions are consecutive integers. y) = (−3. and then we morphed to get y as a function of x in both cases.
Enter the Cubic
The third time around in Chap. the real solutions of the two-by-two system vanish. it makes sense to choose consecutive integers on either side of them. −2) (x. and two larger than −1. we solved this two-by-two system: x 3 + 6x 2 + 14x − y = −7 and −6x + 2y = 2 As with all the other examples.values less than −3.
Solution
To cause the two solution points to merge. y) = (−2. assuming that the other quadratic function (shown by the solid curve) stays the same.
. 27. and also assuming that the contour of the graph for the modified function stays the same. and the parabolas no longer intersect. We also include two x. giving us y = −3x 2 − 5x − 5 If we reduce the constant to anything smaller than −5. along with the results of plugging those values into the functions and calculating. All three solutions are included. y) = (−1. −8)
First. we let x be the independent variable. we must move the dashed parabola downward to the same extent we moved the solid parabola upward in the previous example. That means we must reduce the constant in the function for the dashed parabola by 16.

With six divisions going out from 0 to the left and six to the right. we plot the solution(s) By examining Table 28-3.–2)
Figure 28-3
Graphs of y = x3 + 6x2 + 14x + 7 and y = 3x + 1. each increment is 5 units. each increment is 1 unit. and each increment on the y axis represent 5 units. The first function is graphed as a solid curve. Selected values for graphing the functions y = x 3 + 6x 2 + 14x + 7 and y = 3x + 1. that gives us an absolutevalue span from 0 to 6 for x.
. we can see that the span of absolute values for the input is from 0 to 5.472 More Two-by-Two Graphs
Table 28-3. and that’s enough. For y.
y
(–2. while the absolute values of the functions go up to 38. we can plot the three solution points as shown in Fig. Realnumber solutions appear as points where the curve and the line intersect. we have six divisions going up and eight going down. the second function is graphed as a dashed line.–8)
x
(–1.–5) (–3. On the x axis. and that’s sufficient to include all the function values in Table 28-3. 28-3. Bold entries indicate real solutions. Now that we’ve decided on the dimensions of the coordinate grid. On the y axis. This time. let’s make each increment on the x axis represent 1 unit.
x −5 −4 −3 −2 −1 0 1 x 3 + 6x 2 + 14x + 7 −38 −17 −8 −5 −2 7 28 3x + 1 −14 −11 −8 −5 −2 1 4
Next.

and that’s a span
. You can also include more points “farther out. The solution is in the middle. cubic. the constant. and the approximate graph for y = 3x + 1 is the dashed line. 95/27).Enter the Cubic
473
Finally. we solved the following two cubic functions as a two-by-two system: y = 5x 3 + 3x 2 + 5x + 7 and y = 2x 3 + x 2 + 2x + 5 We got one real solution. 27. that gives us a span from −3 to 3 for x.
Here’s a challenge!
In the “challenge” at the end of Chap. written in bold. A programmable calculator. 28-3. (x. y) = (−2/3. or even 1/10 unit for x-values between −4 and 0 or between −5 and 1. we plot the rest We can fill in the graphs by plotting the remaining points in the table. or higher-degree function based on coefficients. the approximate graph for
y = x 3 + 6x 2 + 14x + 7 is the solid curve. Let’s make each increment on the x axis represent 1/2 unit. and can save you from having to do a lot of tedious arithmetic.
Solution
Table 28-4 shows several values of x. For y. or a personal computer with calculating software installed. In Fig. With six divisions going out from 0 to the left and six to the right. You might also find a site on the Internet that can calculate values of a linear.” say for x-values of −7. 10. along with the resulting function values. If you want to get a “finer” graph in that region. while the span of values of the functions is from −116 to 69. and two complex-conjugate solutions. Draw a graph showing these two functions. The span of values for the input is from −3 to 2. and 15 on the positive side. you can plot points at intervals of 1/2 unit. and −15 on the negative side and 5. quadratic. and input values you choose. we have eight divisions going up and 12 divisions going down. −10.
Are you confused?
Figure 28-3 doesn’t show the relationship between the curve and the line very well in the vicinity of the solution points. 1/5 unit. makes an excellent assistant for this process. along with the real solution point. and each increment on the y axis represent 10 units.

Once we’ve plotted it. But don’t spend too much time at this. and connect the points with smooth curves. If you want to get familiar with how the graphs of various cubic functions look. The bold entry indicates the real solution.52 5 10 29
of −120 to 80. perhaps flipped over backward.
x −3 −2 −1 −2/3 Approx.
. as shown in Fig. which has a limited range with an absolute maximum or an absolute minimum. 3.52 7 20 69 2x 3 + x 2 + 2x + 5 −46 −11 2 95/27 Approx. The graph of a cubic function also has something else that you’ll never see in the graph of a quadratic: an inflection point. and the approximate graph for y = 2x 3 + x 2 + 2x + 5 is the dashed curve. a cubic function always has a range that spans the entire set of real numbers. They all look rather like distorted images of the letter “S” tipped on its side. 28-3 and 28-4? They’re a lot different from the graphs of quadratics! The graph of a cubic function always has one of six characteristic shapes. although it can have a local maximum and a local minimum. The contour of the graph depends on the signs and values of the function’s coefficients and constant.52) This point is shown as a solid dot in Fig. and then extended forever upward and down. we fill in the graphs of the functions. −0. Selected values for graphing the functions y = 5x 3 + 3x 2 + 5x + 7 and y = 2x 3 + x 2 + 2x + 5. 28-5. Then we get (x. 3. you can conjure up a few cubic functions with assorted coefficients and constants. y) = (−0. Unlike a quadratic function. A book devoted to the art of graphing cubic and higher-degree functions could consume thousands of pages! You’ll learn more about graphing functions when you take a course in calculus. Then plot a couple of dozen points for each function.
Are you still confused?
Do you wonder about the “cubic curves” in Figs.474 More Two-by-Two Graphs
Table 28-4. 28-4.67 0 1 2 5x 3 + 3x 2 + 5x + 7 −116 −31 0 95/27 Approx. The approximate graph for y = 5x 3 + 3x 2 + 5x + 7 is the solid curve. where the curvature reverses direction. To plot the solution point. more than enough to include all the function values in Table 28-4. 3.67. we can convert the values to decimal form and go to a couple of decimal places.

Draw the second function’s graph as a dashed line or curve. The first function is graphed as a solid curve. Create a table for both functions based on x-values of −3. C.
y
(–2/3. −2. 1. Plot an approximate graph showing the curves based on the table you created when you worked out Prob. On the x axis.
. the second function is graphed as a dashed curve. if any exist. Don’t hurry! You’ll find worked-out answers in App. Plot and label all real solution points. each increment is 10 units. and 2.95/27) x
Practice Exercises
This is an open-book quiz. −1. The solutions in the appendix may not represent the only way a problem can be figured out. 0. 1. by all means try it! 1. Draw the first function’s graph as a solid line or curve. if any exist. let each increment represent 3 units. let each increment represent 1/2 unit.Practice Exercises
475
Figure 28-4
Graphs of y = 5x3 + 3x2 + 5x + 7 and y = 2x3 + x2 + 2x + 5. If you think you can solve a particular problem in a quicker or better way than you see there. On the x axis. On the y axis. Look again at Practice Exercise 1 and its solution from Chap. The realnumber solution appears as the point where the curves intersect. 27. You may (and should) refer to the text as you solve these problems. −1/2. On the y axis. 2. −3/2. Here are the functions that came from the original equations: y = −3x + 1 and y = 2x 2 + 1 Use bold numerals to indicate the real solutions. each increment is 1/2 unit.

based on x-values of −3. −1. Look again at Practice Exercise 3 and its solution from Chap.476 More Two-by-Two Graphs
A
B
C
D
Local max. Local min.”
3. At E and F. F Local min. with inflection points in the “backward zones. the curves maintain upward or downward trends. −2. 0.
Local max. and 4. Here are the functions that came from the original equations: y = −3x + 1
. Create a table of values for both functions. the curves trend mainly upward or downward but “back up” between local maxima and local minima. At C and D. 3. 2.
E
Figure 28-5 Characteristic shapes for graphs of cubic functions. although they level off at the inflection points (solid dots). 27. At A
and B. and they don’t level off at the inflection points. the curves maintain upward or downward trends. 1.

if any exist. if any exist. Here are the functions that came from the original equations: y = x2 + x + 1 and y = x 2 − 2x − 2 Use bold numerals to indicate the real solutions. Draw the first function’s graph as a solid line or curve. On the x axis. Look again at Practice Exercise 7 and its solution from Chap. 8. 1. based on x-values of −3. Plot and label all real solution points. Look again at Practice Exercise 5 and its solution from Chap. −2. 2. 3. 6. 3. 27. Plot and label all real solution points. let each increment represent 10 units. if any exist. On the y axis. Plot an approximate graph showing the curves based on the table you created when you worked out Prob. 1. if any exist. −2. 0. 7. let each increment represent 1/2 unit. 5. let each increment represent 1 unit. if any exist. Plot an approximate graph showing the curves based on the table you created when you worked out Prob. On the x axis. −3. 0. Create a table of values for both functions. let each increment represent 5 units. On the x axis. −1. let each increment represent 1 unit. 7. Plot an approximate graph showing the curves based on the table you created when you worked out Prob. 5. if any exist. Draw the second function’s graph as a dashed line or curve. 2.
. 4. Draw the second function’s graph as a dashed line or curve. Draw the first function’s graph as a solid line or curve. and 3. 27.Practice Exercises
477
and y = 2x 2 − 3x + 3 Use bold numerals to indicate the real solutions. Here are the functions that came from the original equations: y = −x 2 and y = 2x 3 Use bold numerals to indicate the real solutions. Draw the second function’s graph as a dashed line or curve. On the y axis. let each increment represent 4 units. based on x-values of −4. and 4. Draw the first function’s graph as a solid line or curve. Plot and label all real solution points. On the y axis. Create a table of values for both functions. −1.

Plot an approximate graph showing the curves based on the table you created when you worked out Prob. let each increment represent 1 unit. 1. Create a table of values for both functions. based on x-values of −3. 3. if any exist. On the x axis. −2. 10. 2. Here are the functions that came from the original equations: y = 2x 3 + x 2 + x − 4 and y = 3x 3 − 2x 2 + 4x − 5 Use bold numerals to indicate the real solutions. Draw the first function’s graph as a solid line or curve.478 More Two-by-Two Graphs
9. Plot and label all real solution points. −1. 27. Draw the second function’s graph as a dashed line or curve. On the y axis.
. and 4. 9. Look again at Practice Exercise 9 and its solution from Chap. if any exist. 0. let each increment represent 20 units.

the natural-log function is usually denoted by writing “ln” or “loge” followed by the argument.480 Logarithms and Exponentials
We can raise negative numbers to real-number powers.1 = −1 log10 0. Here are some equations using the natural-log function.5 ≈ −0.807 ln 10 ≈ 2. We aren’t likely to encounter any “base-(−10)” or “base-(−e)” logarithms. although technically there is no reason why such things can’t exist.”
Natural logs Base-e logarithms are also called natural logs or Napierian logs.155 log10 0. which you can check out with your calculator:
ln 100 ≈ 4.303 ln 6 ≈ 1. called the argument. for example.7782 log10 1 = 0 log10 0. for which you want to find the logarithm. Here are a few examples that you can verify with your calculator:
log10 100 = 2 log10 45 ≈ 1.
Common logs Base-10 logarithms are also known as common logarithms or common logs.” The second equation is another way of writing 101. and books.653 log10 10 = 1 log10 6 ≈ 0.605 ln 45 ≈ 3.792 ln e = 1 ln 1 = 0
.07 ≈ −1.01 = −2 The squiggly equals sign means “is approximately equal to.653 ≈ 45 You could also say “The common log of 45 is approximately equal to 1.653. In equations.” You’ll often see it in scientific and engineering papers. and then the number. but this is rarely done with logarithmic functions. The first equation above is another way of writing 102 = 100 You could also say “The common log of 100 is equal to 2. articles. common logs are denoted by writing “log” with a subscript 10. In equations.3010 log10 0.

always include the base as a subscript whenever you write “log” followed by an argument.807 ≈ 45 You could also say “The natural log of 45 is approximately equal to 3. then it’s the common log.6931 ln 0.807. If the “log” of 10 equals 1.
Solution
Note that 0.1 ≈ −2.3. followed by the argument.07 ≈ −2. e. You can have logs of arguments that are irrational. and 100 = 102. 10. But in other texts and in most calculators. fractions. Therefore: log10 0.659 ln 0.303 ln 0.What Is a Logarithm?
481
ln (1/e) = −1 ln 0. “log” means the common (base-10) log.1 = −1 log10 1 = 0 log10 10 = 1 log10 100 = 2
. base-e logs) are denoted by writing “log” without a subscript. then it’s the natural log.605.1 = 10−1. In some texts.605 The first equation above is another way of writing e4. Then compare the natural logarithms of those same arguments.” The second equation is an alternative way writing e3. such as π. natural logs (that is.1. 0. 0. 1 = 100. For example. or the cube root of 100.
Here’s a challenge!
Compare the common logarithms of 0. and 100. If the “log” of 10 equals an irrational number slightly larger than 2. To avoid confusion. 10 = 101. do a trial calculation to find out. the positive square root of 2.605 ≈ 100 You would say it as “The natural log of 100 is equal to approximately 4.01 = 10−2.
Don’t let them confuse you!
Authors don’t all agree on what the notation “log” means. You don’t have to use a subscript when you write “ln” for the natural log.01 = −2 log10 0.01 ≈ −4. write “log10” or “loge” instead of “log” all by itself. 1.5 ≈ −0. If you aren’t sure what the “log” key on a calculator does.01. or terminating decimals.” The arguments in common or natural log functions don’t have to be whole numbers.

regardless of the base b.0792 You shouldn’t expect to get perfect answers every time you use logarithms. is always equal to the sum of the logarithms of the individual numbers. Consider the arguments exact. you should get log10 12 ≈ 0. The base-e logarithm of a number is the power of e that produces that number. suppose that x and y are positive real numbers. non-repeating decimals. because any positive real number raised to the zeroth power is equal to 1.6021 ≈ 1.4771 + 0.3026 ln 1 = 0 ln 10 ≈ 2. is equal to the difference between the logarithms of the individual numbers:
logb (x /y) = logb x − logb y
.3026 ln 100 ≈ 4. Use your calculator to follow along: log10 (3 × 4) = log10 3 + log10 4 Working out both sides and approximating the results to four decimal places. rounded off to four decimal places in each case (except the natural log of 1.01 ≈ −4.482 Logarithms and Exponentials
These values are all exact! But now suppose you want to compare the natural logs of these same five arguments.
Changing a ratio to a difference Again. which is an exact value): ln 0. You must use a calculator to find natural logs. You can write this as
logb xy = logb x + logb y Let’s look at a numerical example. no matter what the base.6052 ln 0. The results are as follows.1 ≈ −2. That means they are endless. Approximation is the best you can do.
Changing a product to a sum Imagine two positive real number variables x and y. you can convert products into sums. powers into products. because the results are almost always irrational numbers. Then the logarithm of their ratio. ratios into differences. The logarithm of their product.6052 The log of 1 is always equal to 0. and roots into ratios.
How Logarithms Work
With logarithms. no matter what the base b happens to be (as long as b > 0).

carried out to four decimal places: log10 (34) = 4 log10 3 log10 81 ≈ 4.How Logarithms Work
483
You can work out an example using the same numerical arguments as before. The base-b logarithm of x raised to the power y can be rearranged as a product:
logb x y = y logb x Again.9084 Now let’s try an example in which both of the numbers in the input argument are decimals.1250. Let x be a positive real number.6021 ≈ −0. which often occurs when repeated calculations are done using approximate values. This is an example of a phenomenon called rounding error. you’ll see −0. This isn’t the last time you’re going to see it!
Changing a power to a product Logarithms simplify the raising of a number to a power. this works for any positive base b.7500 directly with your calculator and round it off to four decimal places.1249. Again.1250
Are you confused?
If you take the base-10 log of 0.
. not −0. We’ll go to three places: log10 (2.4771 − 0. This tactic is useful when the argument does not have two whole numbers.078 log10 2. Here’s an example using the same arguments as before.4771 ≈ 1. What’s going on? Why is there a discrepancy between these two methods of determining the base-10 log of 3/4? It’s not a flaw in the calculator.421 ≈ 0.635 ≈ 1.078) = 1.6351.0000 × 0. follow along with your calculator: log10 (3 /4) = log10 3 − log10 4 Working out both sides and approximating the results to four decimal places: log10 0.078 × 0.454 We’ll return to this result later. and it’s not your imagination. and it’s especially handy when irrationals enter the scene.7500 ≈ 0. and let y be any real number.

we can evaluate both expressions. and using natural logs evaluated to five decimal places. Then the logarithm (to any positive base b) of the yth root of x (also denoted as x to the 1/y power) is equal to the log of x. We get ln (81/3) = (ln 8) / 3 We know that the 1/3 power (or cube root) of 8 is equal to 2.07944 / 3 0.333333333 ≈ −(ln 3. ln 0. This time.000!
Reciprocal within an exponent What happens when we have a reciprocal in an exponent? Suppose x is a positive real number. especially if you get into physics or engineering.
. Stated mathematically. then
logb (1/x) = −(logb x) This is a special case of division simplifying to subtraction. Suppose x = 3 (exactly) and we use natural logs.69315 The results agree to all five decimal places here. considering both values exact.
Log conversions Here are a couple of useful rules for converting natural logs to common logs and vice-versa. and y is any real number except zero. if x is a positive real number.000000000) −1. as follows: ln (1/3) = −(ln 3) Using a calculator.000. You’ll sometimes have to do this. Therefore ln 2 = (ln 8) / 3 0. but it’s only at the ninth decimal place.098612289 We have some rounding error here. divided by y :
logb (x 1/y) = (logb x) / y Let’s try this with x = 8 and y = 1/3. equivalent to one part in 1.69315 = 0.484 Logarithms and Exponentials
Changing a reciprocal to a negative The logarithm (to any base b) of the reciprocal of a number is equal to the negative of the logarithm of that number.098612290 ≈ −1.69315 ≈ 2.000. let’s go to nine decimal places. so we can write a plain equals sign instead of a squiggly one. Now let’s look at a numerical example.

Again.537 ≈ 1. Suppose x is a positive real number.434294 comes out as 2.
Are you confused?
“Is there any way.302525 when rounded to six decimal places. If we go to six decimal places to approximate the natural log of 10.434294 ≈ 2.434294 × 1.537 ≈ 0. getting 0.537 as a calculator determines it.302588 when you round it off to six decimal places. To demonstrate with a calculator. “to eliminate rounding errors in calculations?” The answer is. but it can be a little tricky.302585 we obtained above.549 Now let’s go the other way. you can carry out your calculations to many more decimal places than necessary until the very end.” you ask. and then—but only then—round the values off to the number of places you want. its reciprocal comes out as 0. the reciprocal of 0.302585 ≈ 0. The common logarithm of x can be expressed in terms of the natural logarithms of x and 10. sometimes. we can use a calculator to find ln 3. Working to three decimal places.How Logarithms Work
485
Suppose that x is a positive real number. “Yes. this is an example of rounding error. The natural logarithm of x can be expressed in terms of the common logarithms of x and e.302588 log10 x If you’re astute. we have ln x = (log10 x) / (log10 e) ≈ (log10 x) / 0. Let x = 3.
. But if you take the constant derived from ln 10 earlier. we have log10 x = (ln x) / (ln 10) ≈ (ln x) / 2. you’ll notice that this value differs slightly from the constant 2.537. again rounding off to three decimal places: log10 3.434294.434294 ln x Let’s try a numerical example. which is 2.434294 and round the answer off to three decimal places.263 We multiply by 0. or any other numbers where the values can be approximated but never exactly written down.263 ≈ 0.549 We can compare this with the common log of 3.” When working with irrational numbers. If we go to six decimal places to approximate the common log of e.

When you get the final answer and round it off to.967 ≈ 5. say. It’s impossible to find any real number x. you can leave all the displayed digits in the figures as you go through the arithmetic. You can play around with a good calculator to see how this works.378 ≈ 5. those extra digits can be kept in the process. check out what happens if you try to calculate the common log of −2.476 When you take the natural log of 238. but it’s not a real number! No matter what real number you choose for x. “These are not defined in the set of real numbers. Then take the natural log of 238. six decimal places instead of the two or three dozen. Suppose that log10 −2 = x This can be rewritten as 10x = −2 What’s the value of x ? That’s hard to say.378.302588 log10 x In this example.486 Logarithms and Exponentials
If you have a high-end calculator that goes to two or three dozen digits such as the sort found in some computers. round it off to three decimal places.302588 by 2.378 Now remember the conversion formula: ln x ≈ 2.
Here’s a challenge!
Find the natural log of 238.967 directly with a calculator and then round off to three decimal places.” To understand why.302588 × 2.476
. you’ll see that ln 238.
Are you still confused?
You might also ask.967 ≈ 2. and compare that result with the first one. and that will keep the rounding error extremely small.967 from its common log using the above conversion formula. “What about the logarithm of 0. the error won’t show up. no matter what the base. Express your answer to three decimal places. such that 10x is not a positive real. you run into the same problem. When you round the answer off to three decimal places. you should multiply 2. or of any negative number?” The answer is.967 directly with your calculator. or to 0. Using the memory store and recall functions. the value of 10x is positive. you should get 2.
Solution
Working with a calculator. If you change −2 to any other negative number. you should get log10 238.

1 10−1.
Here’s a twist!
It’s possible to raise imaginary numbers to real-number powers and get negative numbers. Consider this: j 2 = −1 You won’t be likely to hear anybody state this fact as “Minus 1 is the base-j exponential of 2.020 10−2 = 0. you’ve already learned about this.5 ≈ 0. is raised to a variable power. e. it’s often called the exponential constant.” But theoretically.834 100 = 1 10−0.316 10−1 = 0. and you raise b to the xth power to get y. it’s a valid statement. x.0855 is the exponential.061 101 = 10 100.7 ≈ 0. rounding to three decimal places except in those cases when the resultants are exact:
102 = 100 101. When Euler’s constant.What Is an Exponential?
487
What Is an Exponential?
The exponential of a quantity is what you get when you raise a certain positive real number. In the expression e3 ≈ 20. In fact. like this:
y = bx Then y is the base b exponential of x.8347 ≈ 6.01
. just as a logarithm has.
Common exponentials Base-10 exponentials are also known as common exponentials. and y. Here are a few examples that you can verify with your calculator.0855. where b > 0. As is the case with logarithms. called the base. to a power equal to that quantity.478 ≈ 30. In the expression 102 = 100. e is the base and 20. for example. You could talk about base-j exponentials. an exponential function is an “inside-out” way of looking at a logarithmic function! Suppose you have three real numbers b.
The exponential base An exponential function has a base and an exponent. 10 is the base and 100 is the exponential. and make mathematical sense. the two bases you’ll most often see are 10 and e.

the arguments in an exponential function need not be whole numbers. and 10(log y) = y for any positive real number y. you could speak of 10π. Here are some examples using the same arguments as above. The natural exponential of a quantity is sometimes called the natural antilogarithm (antiln) or the natural inverse logarithm (ln−1) of that number. they “undo” each other. They can even be irrational.)
Natural exponentials Base-e exponentials are also called natural exponentials. You would say is as “The common exponential of 2 is equal to 100. The reverse of this is also true: an exponential can be “undone” by the log function of the same base.478
Logarithms vs.478 is approximately 30. which is exactly 1:
e 2 ≈ 7. and 100 is the resultant.135
1.7 ≈ 0.183 e−2 ≈ 0.304 0 e =1 e−0. for example. We can illustrate the relationship between a common log and a common exponential with two equations. If we let the abbreviation “log” represent the base-10 logarithm. you would say “The common exponential of 1.384 1 e ≈ 2. the common exponential of a quantity is called the common antilogarithm (antilog10) or the common inverse logarithm (log−1) of that number.8347 e ≈ 2. then
log (10x) = x for any real number x. (It’s approximately equal to 1. and vice-versa.061. When two functions are inverses. A similar pair of equations holds for the natural logarithms. Sometimes.368 e−1.607 e−1 ≈ 0. 2 is the argument (the value on which the function depends). We can replace 10 with e. A logarithm can be “undone” by the exponential function of the same base. as long as both functions are defined for the all the arguments of interest. exponentials The exponential function is the inverse of the logarithm function.385.” In the second equation.488 Logarithms and Exponentials
In the first equation. rounding to three decimal places except for e0.389 e ≈ 4. and replace “log” with “ln” to get ln (e x) = x
.” As with logarithms.718 0.5 ≈ 0.

078 ≈ 2.6351.14159.078 log10 2. and e(ln y) = y for any positive real number y. Recall the example set out earlier in this chapter:
log10 (2. The log-antilog scheme is the way most calculators work out powers when the input values are not whole numbers. and simplify the left side of the equation based on the fact that the antilog function “undoes” the log function.078)] ≈ antilog10 0.454 We can use a calculator to find the common antilog of 0.078) ≈ 0.002 (or 2 parts in 1.078 were mysterious. expressions such as 2.6351. introducing a rounding error. as long as the log and the antilog are both defined for all the arguments.078 ≈ 2.6351. we have log10 (2. We can use logs and antilogs of any base to evaluate any number raised to the power of any other number. When I enter the original numbers into my calculator and use that key. I get 2.What Is an Exponential?
489
for any real number x.454 If we get rid of all the intermediate expressions.
An example Now let’s see how a common antilog can be used to find the value of a non-whole number raised to the power of another non-whole number. and is an irrational number equal to approximately 3.078) = 1.421 ≈ 0.
.6351.000) because we rounded off the calculations at every step. Before the invention of logs and antilogs.842 This answer disagrees from the previous answer by 0. we get antilog10 [log10 (2.
Another example What do you get if you raise e to the power of π? You should remember from basic geometry what π (the lowercase Greek letter called “pi”) means.6351.078 × 0.844 We can use the “x ^y” key in a calculator to verify the above result. It’s the ratio of any circle’s circumference to its diameter. When we do that.635 ≈ 1. indeed. we get 2.454 Taking the common antilog of both sides.6351.454.

71828 as an approximation of e. This time. Use the same process as you did to find e π in the example you just finished. and rounding off to five decimal places when you get to the very end. the above equation becomes log10 e π ≈ 3. because antilog10 (log10 e π) = e π Calculating.14159 as an approximation of π.71828 ≈ 3.14159 × 0.14159 ≈ 23. you can repeat this exercise. you should get antilog10 1.14159 log10 2. Round off the values to five decimal places.
Here’s a challenge!
Using a calculator. but it does have a common log key. and because e is a little less than 3. going to five decimal places in each step.490 Logarithms and Exponentials
Suppose you have a calculator that can’t directly raise one number to the power of another. the value of e π should be somewhere near 33. When I input the numbers into my calculator. you can check this out. You can calculate e π using the rule for converting a power to a product: log10 e π = π log10 e Take the value 2. Think of it like this: Because π is a little more than 3.14058 As usual. but it’s “in the ball park”! If your calculator has a key for raising one number to the power of another (in general).36436 When you take the common antilog of this.
Solution
Once again.36436 ≈ 23. Then. you have log10 π e = e log10 π
. rounding error has crept into this process. It’s not terribly close. you can take advantage of the rule for converting a power to a product.13982 This makes intuitive sense. you’ll get the value of e π. I get 2.718283. and 3. letting your calculator keep all the extra digits it can along the way. which is 27.43429 ≈ 1. If you like. find πe. Verify this result by using the “x ^y” key if your calculator has one.

How Exponentials Work
In most real-life applications of exponentials. if you wish. This is the same as saying that 1/8 is equal to 2−3.49715 ≈ 1.71828 × 0. The reciprocal of the exponential of x is equal to the exponential of the negative of x.71828.71828 ≈ 22.45898 When I input the numbers into my calculator and use the “x ^y” key. Then log10 πe ≈ 2. You know that 1/8 is equal to 1 / (23).71828 log10 3.45906
491
Rounding error strikes again! And as before. Now consider this. This section describes the basic properties that hold for exponentials in general. You also know that 1/100 equals 1 / (102).0498
. once in a while you’ll come across a situation where b is some other positive real number. you’ll get πe because antilog10 (log10 πe) = πe Calculating.
Reciprocal vs.7183) ≈ 1 / 20.35139 ≈ 22. negative exponent Suppose that x is some real number. However. which is the same as saying that 1/100 equals 10−2.14159. the base b is either 10 or e. as follows:
1 / (b x) = b −x when b > 0. and consider e equal to 2.35139 When you take the common antilog of this. you should get antilog10 1. You should recognize this from your work with powers and roots. rounded to four decimal places: 1 / (e 3) ≈ 1 / (2.079 ≈ 0. I get 3. Here’s a familiar example. you can see for yourself how this error shrinks to the vanishing point if you let your calculator keep all of its extra digits until the final step.14159 ≈ 2.How Exponentials Work
Consider π equal to 3.141592.

000 / 0. We plug in the numbers on the left side of the equation and get 104 / 10−6 = 10. you should expect some rounding error.000001 = 0.000 = 10. no matter what base and arguments you use.” then hitting “ln.000 × 0. sum Exponential functions can express the relationship between sums and products. difference Again.000. Then
b xb y = b (x+y) when b > 0.000. We can plug in the numbers for the product of exponentials and get and get 104 × 10 −6 = 10. we can demonstrate this. Of course. Using the same numerical values as before. then hitting “Inv. if you get nonterminating decimals for any of the values in the calculation.000001 = 10.000 × 1.000. let b = 10.492 Logarithms and Exponentials
Compare the above with the result of entering −3 into a scientific calculator. suppose that x and y are real numbers.01 The results agree.01 When we evaluate the right side. as long as the base is positive. x = 4. we get 10[4+(−6)] = 10(4−6) = 10−2 = 0.” and finally rounding to four decimal places: e−3 ≈ 0. Suppose that x and y are real numbers. Then
b x / b y = b(x−y) when b > 0.
Ratio vs.000 = 1010 Then we can evaluate the right side to see that 10[4−(−6)] = 10(4+6) = 1010
. You’ll find that this is always true. To demonstrate. just as logarithms do. and y = −6.0498 Rounding error stayed out of this little exercise!
Product vs.

The result is 3. let b = e. with exponents x = 4 and y = 7. Now when we plug the numbers into the right side of the general equation and work it out.000(1/7) To work this out on a calculator. Let x and y be real numbers. hit the “Inv” key.1429. with the restriction that y cannot be equal to 0.1429. Then. and enter 0. we can enter 0. Then
(b x)y = b(xy) when b > 0. we obtain (104)(1/7) = 10.729.727 Alternatively. To demonstrate this.3883 ≈ 403. and then hit “log” to find 10 to the power of 0.5714. Let’s try an example where the base b is 10.256 Now the right side: e(2×3) = e6 ≈ 2. we enter 10. we get 10(4/7) ≈ 100. product Exponentials can show the relationship between a “power of a power” and a product.000. and y = 3.7186 ≈ 403. x = 2. hit the “x y” or “x ^y” key. Evaluating the left side first.5714 ≈ 3. That gives us 0. There’s a discrepancy. Then
b(x /y) = (b x)(1/y) when b > 0. because we’ve taken a rounding error to the seventh power!
Power of a power vs. we must first figure 1/7 to four decimal places.178
.718 as the value of e and going to three decimal places during the calculation process: (e 2)3 ≈ (2.How Exponentials Work
493
Ratio in an exponent Here’s a more complicated property of exponentials. using 2.5714. Suppose that x and y are real numbers.7182)3 ≈ 7. Let’s evaluate the left side first.5714 and using the “x y” or “x ^y” function key on a calculator. letting 4/7 ≈ 0.

Expressing the value of e to five decimal places but rounding off our final answer to only two decimal places.718284 ≈ 10.000 = 106 and the right side becomes 10(2 ×3) = 106 There’s no rounding error here. If x is the argument of both exponentials.18284 ≈ 545. 9.16
. the left side of the above equation works out as (104)(e4) ≈ 10.) Now let’s use b = 10 instead of b = e. Let x = 4. Let’s try a numerical example.980 and the right side becomes (10e)4 ≈ (10 × 2.59800 ≈ 183. as in the previous example. because the values are exact throughout!
Mixing exponentials in a product We can express the product of a common exponential and a natural exponential a “mutant exponential” whose base is 10 times e. the left side becomes (102)3 = 1003 = 1.000 / 2. 9.71828)4 ≈ 27. In that case. we don’t get the rounding error when we go back to three decimal places at the end.718284 ≈ 10.000.000 × 2.494 Logarithms and Exponentials
If we let the calculator keep all its extra digits during the calculation process.980
Mixing exponentials in a ratio How about ratios of mixed common and natural exponentials? If x is a real number. the left side of the above equation becomes 104 / e4 ≈ 10.000 × 54. then
(10 x)(e x) = (10e)x This is an adaptation of the power of product rule from Chap. (Try it and see. If we express the value of e to five decimal places. then
10x / e x = (10/e)x This is an adaptation of the power of quotient rule from Chap.000 / 54.59800 ≈ 545. We can work this out using x = 4.

0. the base is 1!) Then we do the same thing with the bases 2.678804 ≈ 183.000
. we can solve the following equation for x : e x = 1. First. all the results are exact. −1. we must be sure that we know what we’re trying to get! Suppose we call the solution x. For the bases 1.718284 / 10. Finally. here’s a trick to get the general idea. For the base e.000. the left side of the above equation works out as e4 / 104 ≈ 2.0001. Then we can write e x / 10 x = (e /10)x Again. (That’s right. 2. 2. Expressing the value of e to five decimal places and leaving our final answer at five decimal places too. and 10.00546
Are you confused?
To get a “snapshot” of how exponentials of different bases work out. we determine the values of 1 to the powers of −4.000.000.How Exponentials Work
495
and the right side becomes (10/e)4 ≈ (10 / 2.000 ≈ 54. −2.
Solution
To solve this problem. we approximate everything to four decimal places except e 0. e. In the first case.2718284 ≈ 0. we compare the values as shown in Table 29-1. and 10. −3.00546 and the right side works out as (e /10)4 ≈ (2. 1. let’s use x = 4 and go through this with a calculator.71828 / 10)4 ≈ 0.71828)4 ≈ 3. Suppose x is a real number. and the number whose natural exponential function value is exactly 0.16 Now let’s invert this ratio.000 ≈ 0.
Here’s a challenge!
Find the number whose natural exponential function value is exactly 1.59800 / 10. which is exactly 1. 3. and 4.

125 0.82 In the second case.000 ≈ 13. Rounded off to two decimal places.25 0.000 This simplifies to a matter of finding a natural logarithm with a calculator.000 10. and then round off to two decimal places.001 0.1 1 10 100 1.21
Practice Exercises
This is an open-book quiz.0183 0. we get ln (e x) = ln 1. we get ln 0.0001 This simplifies. by all means try it!
.7183 7. Don’t hurry! You’ll find worked-out answers in App.3891 20.0625 0. we have ln 1.000.
1x 1 1 1 1 1 1 1 1 1 2x 0.0498 0.0001 0. we get ln (e x ) = ln 0.000 x = ln 1. You may (and should) refer to the text as you solve these problems.1353 0. which is exactly 1.0001 Taking the natural logarithm of each side.0001 ≈ −9.5982 10x 0. The solutions in the appendix may not represent the only way a problem can be figured out. e. to a matter of finding a natural logarithm with a calculator.0855 54. 2.000.0001 x = ln 0.3679 1 2.496 Logarithms and Exponentials
Table 29-1. and 10. we must solve the following equation for x : e x = 0.000.
x −4 −3 −2 −1 0 1 2 3 4
Comparison of exponentials for bases 1. If you think you can solve a particular problem in a quicker or better way than you see there. When we do that.01 0.5 1 2 4 8 16 ex 0. Values for base e are approximate except for e0. as in the first case.000
Taking the natural logarithm of each side. C.

If a positive real number is divided by a factor of 357. 3. What is the power gain of the length of speaker wire.7 watts. Show that the solution to Prob. 7.902780337. Approximate the product xy from Prob. 5. What is the power gain of this circuit in decibels? Round off the answer to the nearest tenth of a decibel. the speaker only gets 19. can be calculated according to this formula: G = 10 log (Pout/Pin) where G is the gain. 1 using natural logarithms. how does its natural (base-e) logarithm change? Express the answer to two decimal places. how does its common (base-10) logarithm change? 6. Suppose the audio output signal in the scenario of Prob. in units called decibels (abbreviated dB).3 watts. how does its common logarithm change? 8. Find xy to three decimal places using common logarithms. in decibels? Round off the answer to three decimal places. 7 is valid for all positive real numbers. and Pin is the input signal power. both specified in watts. Show that the result is the same as that obtained with common logs when rounded off to three decimal places.
. 10. 9. If a positive real number decreases by a factor of exactly 100 (becomes 1/100 as great). Show that the solution to Prob. 9 is valid for all positive real numbers.3713018568 and y = 0. If a positive real number increases by a factor of exactly 10. and the output is 23. so instead of the 23. Show that the solution to Prob.535 watts.7 watts that appears at the left-channel amplifier output. Suppose the audio input to the left channel of high-fidelity amplifier is 0. The power gain of an electronic circuit. 2.Practice Exercises
497
1. 5 is valid for all positive real numbers. Pout is the output signal power. 4. 3 is run through a long length of speaker wire. Let x = 2.

(We always put the minus sign first. If b is positive. then we write the product of j and b as jb.Part Three 499
The power of product rule allows us to rewrite the left side of this equation. we would more likely write j0 simply as 0. using letter constants instead of numbers! The distributive law tells us that ja + jb = j(a + b)
. so it becomes [(−1)j ]2 = −1 But (−1)j is the same thing as −j. then ja + jb is the same as jb + ja?
Answer 21-5
We can use the same proof procedure as we did in Answer 21-4. again assuming that the distributive law works with the unit imaginary number.) If b = 0. But because j0 = 0. Suppose we call the real number b. then we can write the product as j0. If b is negative. we get j3 + j 7 = j(7 + 3) Applying the distributive law “forward” on the right side. that if a and b are any two real numbers. so we can simplify further to get (−j )2 = −1 This is the same result as we obtain by squaring j. assuming that the familiar distributive law of arithmetic works with the unit imaginary number?
Answer 21-4
When we apply the distributive law for multiplication over addition “backward” to the first expression. we get j3 + j 7 = j 7 + j3
Question 21-5
How can we show. we get j 3 + j 7 = j(3 + 7) The commutative law for addition tells us that 3 + 7 = 7 + 3. we write the product as −jb.
Question 21-3
How is an imaginary number “put together”?
Answer 21-3
An imaginary number is the product of j and a real number.
Question 21-4
How can we show that j 3 + j 7 is the same as j 7 + j 3. By substitution on the right side.

or 0.
Question 21-8
What is the sum of the complex numbers a1 + jb1 and a2 + jb2. a2. b1. Conversely. b1. There are no restrictions on the values of a or b. any complex number can be written in the form a + jb. a2. gives us ja + jb = jb + ja
Question 21-6
What is the absolute value of j 25? What is the absolute value of −j 25?
Answer 21-6
Remember that in general. if b is any nonnegative real number. and b2 are real numbers?
Answer 21-8
When we want to add two complex numbers. then a + jb is a complex number. and b2 are real numbers?
.500 Review Questions and Answers
The commutative law allows us to rewrite this as ja + jb = j(b + a) and the distributive law. positive. where a1. If we have two real numbers a and b. They can be negative.
Question 21-7
How is a complex number “put together”?
Answer 21-7
A complex number is the sum of a real number and an imaginary number. where a1. then |jb| = b and |−jb| = b The absolute value of j 25 and the absolute value of −j 25 are therefore both equal to 25. where a and b are real numbers. we add the real parts and the complex parts separately. applied again. (a1 + jb1) + (a2 + jb2) = (a1 + a2) + (jb1 + jb2) = (a1 + a2) + j(b1 + b2)
Question 21-9
What is the product of the complex numbers a1 + jb1 and a2 + jb2. Therefore.

keeping in mind the fact that j 2 = −1. (a1 + jb1)(a2 + jb2) = a1a2 + ja1b2 + jb1a2 + j 2b1b2 = a1a2 + ja1b2 + jb1a2 − b1b2 = a1a2 − b1b2 + ja1b2 + jb1a2 = (a1a2 − b1b2) + j(a1b2 + b1a2)
Question 21-10
What is the conjugate of a complex number a + jb. Therefore. Therefore.Part Three 501 Answer 21-9
When we want to multiply one complex number by another. where a and b are real numbers? What happens when we add a complex number to its conjugate? What happens when we multiply a complex number by its conjugate?
Answer 21-10
We can get the conjugate of any complex number a + jb by reversing the sign of the imaginary part. a + jb and a − jb are conjugates of each other. When we add a complex number to its conjugate using the rule from Answer 21-8. we get (a + jb)(a − jb) = a2 − jab + jba − j 2b2 = a2 − jab + jab + b2 = a2 + b2
Chapter 22
Question 22-1
What is the polynomial standard form for a quadratic equation in the variable x?
Answer 22-1
When a quadratic equation in x is written in polynomial standard form. it’s formatted like this: ax 2 + bx + c = 0
. we get (a + jb) + (a − jb) = (a + a) + (jb − jb) = 2a + j0 = 2a When we multiply a complex number by its conjugate using the rule from Answer 21-9. we treat both factors as binomials.

Question 22-4
What are the roots of the following quadratic equation. and x is the variable? Assume that neither a1 nor b1 is equal to 0: (a1x + b1)(a2x + b2) = 0
. provided a1 ≠ 0 and a2 ≠ 0. a2. and b2 are real numbers. The coefficient b from the “template” is the quantity (a1b2 + b1a2). When we do that. and b2 are real numbers. 9. There may also be terms containing the variable itself (to the first power) along with terms that are simple constants. we get a1a2x 2 + a1b2x + b1a2x + b1b2 = 0 which can be rearranged to obtain (a1a2)x 2 + (a1b2 + b1a2)x + b1b2 = 0 This equation is in polynomial standard form. third.
Question 22-3
Which of the following equations are quadratics in one variable? Which are not? x 2 = 8x + 3x 3 −3x = 7x 2 − 12 x 2 + 2x = 7 − x 4 x − 2 = − 8x 2 − 7x 3 13 + 3x = 12x 2
Answer 22-3
A quadratic equation in one variable always contains a nonzero multiple of the variable squared.
Question 22-2
How can we write the following equation as a quadratic in polynomial standard form when a1. and fifth equations are quadratics in one variable. b1. The coefficient a from the “template” (Answer 22-1) is the quantity a1a2. b1. The constant c from the “template” is the quantity b1b2. and x is the variable? (a1x + b1)(a2x + b2) = 0
Answer 22-2
We can start by multiplying the two binomials on the left side together. using the product of sums rule from Chap. and no higher powers of the variable. The first and fourth equations are not. the coefficient of x 2 (in this case a) must not be equal to 0. For the equation to be a quadratic. where a1.502 Review Questions and Answers
where a and b are coefficients of the variable x. the second. and c is a constant. On that basis. a2.

We can find those roots by setting each binomial equal to 0. and x is the variable? Assume that a ≠ 0: (ax + b)2 = 0
Answer 22-5
We can look at this equation as a product of two identical binomials: (ax + b)(ax + b) = 0 We can solve this quadratic in the same way as we solved the equation in Answer 22-4. For the first term. here’s the general form: ax 2 + bx + c = 0
. and some combination of the coefficient and constant on the right side. Again. −b/a. we get only one root: ax + b = 0 ax = −b x = −b /a
Question 22-6
What’s special about the root of the equation we solved in Answer 22-5?
Answer 22-6
This root.
Question 22-7
Suppose we see a quadratic equation in polynomial standard form.Part Three 503 Answer 22-4
This equation is a product of binomials. the process goes like this: a1x + b1 = 0 a1x = −b1 x = −b1/a1 For the second term. creating two separate first-degree equations. But this time. Then we can morph the equations to get the variable all alone on the left side of the equals sign. The roots are the values of x that make either factor equal to 0. occurs “twice over. but with subscripts of 2 instead of 1: a2x + b2 = 0 a2x = −b2 x = −b2/a2
Question 22-5
How can we solve the following quadratic equation.” In technical terms. where a and b are real numbers. the root has multiplicity 2. it’s the same.

sometimes symbolized as d. We can do that by adding 49 to each side.
.764 − 1. If d > 0. obtaining 9x 2 − 42x + 49 = 0 In the general polynomial standard equation ax 2 + bx + c = 0 we have a = 9. we must get the equation into polynomial standard form. then the equation has no real roots.504 Review Questions and Answers
where a and b are coefficients of the variable x. and c = 49. a ≠ 0. known as the quadratic formula. b = −42. we get x = [−b ± (b2 − 4ac)1/2] / (2a) = [42 ± (422 − 4 × 9 × 49)1/2] / (2 × 9) = [42 ± (1. What is the general formula for solving this quadratic?
Answer 22-7
The formula.
Question 22-9
How can we use the quadratic formula to find the roots of the following equation? 9x 2 − 42x = −49
Answer 22-9
Before applying the formula. If d = 0. It tells us whether or not a quadratic equation with real-number coefficients and a real-number constant has any real roots. is the quantity (b 2 − 4ac).764)1/2] / 18 = 42/18 = 7/3 This equation has the single root x = 7/3 with multiplicity 2. If d < 0. then the equation has one real root with multiplicity 2. Plugging these into the quadratic formula. then the equation has two different real roots. and c is a constant. is x = [−b ± (b 2 − 4ac)1/2] / (2a) This is worth memorizing!
Question 22-8
What is the discriminant in the quadratic formula? Why is it significant?
Answer 22-8
The discriminant.

have any roots at all? If so. both of which are non-real complex numbers.
Question 23-2
How can we use the quadratic formula to find the roots of the following equation? −x = 3x 2 + 4
. We can do that by adding x to both sides and then switching the right and left sides. That gives us 3x 2 + x − 4 = 0 In the general polynomial standard equation ax 2 + bx + c = 0 we have a = 3. but with a negative discriminant. b = 1. the equation has two roots.Part Three 505 Question 22-10
How can we use the quadratic formula to find the roots of the following equation? −x = 3x 2 − 4
Answer 22-10
First. Plugging these into the quadratic formula. we get x = [−b ± (b2 − 4ac)1/2] / (2a) = {−1 ± [12 − 4 × 3 × (−4)]1/2} / (2 × 3) = [−1 ± (1 + 48)1/2] / 6 = (−1 ± 491/2) / 6 = (−1 ± 7) / 6 = 6/6 or −8/6 = 1 or −4/3 The roots of the quadratic equation are x = 1 or x = −4/3. what are they like?
Answer 23-1
When a quadratic equation with real coefficients and a real constant has a negative discriminant. let’s get the equation into polynomial standard form. and c = −4.
Chapter 23
Question 23-1
Does a quadratic equation with real coefficients and a real constant.

we must get the equation into polynomial standard form. How can we tell whether or not these are pure imaginary numbers?
Answer 23-3
If the discriminant is a negative real and the coefficient of x is 0. getting x = −1/6 + j(471/2/6) or x = −1/6 − j(471/2/6)
Question 23-3
Suppose we examine the discriminant of a quadratic equation. then the roots are pure imaginary. then the roots are complex but not pure imaginary. how are those roots related?
. b = 1. and c = 4.
Question 23-4
In a quadratic equation with real coefficients but pure imaginary roots. and we discover that the equation has two complex roots.506 Review Questions and Answers Answer 23-2
First. Therefore x = [−1 ± j(471/2)] / 6 If we want to express these roots individually. Plugging these into the quadratic formula. how are those roots related? In a quadratic equation with real coefficients but complex roots that aren’t pure imaginary. If the discriminant is a negative real and the coefficient of x is a nonzero real. we get x = [−b ± (b2 − 4ac)1/2] / (2a) = [−1 ± (12 − 4 × 3 × 4)1/2] / (2 × 3) = [−1 ± (1 − 48)1/2] / 6 = [−1 ± (−47)1/2] / 6 The quantity (−47)1/2 is the imaginary number j(471/2). getting 3x 2 + x + 4 = 0 In the general polynomial standard equation ax 2 + bx + c = 0 we have a = 3. We can do that by adding x to both sides and then switching the right and left sides. we can write x = [−1 + j(471/2)] / 6 or x = [−1 − j(471/2)] / 6 We can reduce these to standard complex-number form using the right-hand distributive law for division over addition or subtraction.

obtaining (x − j4)(x + j4) = 0 That’s the binomial factor form of the original equation. then they are additive inverses. then they are conjugates. we obtain x 2 = −q /p which can be rewritten as x 2 = −1(q /p)
. one is the negative of the other. That is. We can write the negatives of these roots as constants in a pair of binomials and then set their product equal to 0. If the roots are complex but not pure imaginary. which we know is not 0 because we’ve been told that it’s positive.Part Three 507 Answer 23-4
If the roots are pure imaginary. we get px 2 = −q Dividing through by p.
Question 23-5
Consider the following quadratic equation in polynomial standard form: 4x 2 + 64 = 0 How can this equation be rewritten in binomial factor form?
Answer 23-5
If we subtract 64 from both sides of the equation and then divide through by 4. What are the roots of this equation?
Answer 23-6
Subtracting q from each side. we get x 2 = −16 This tells us that the roots are x = j4 or x = −j4.
Question 23-6
Consider the general quadratic equation px 2 + q = 0 where p and q are both positive real numbers.

provide an example of such an equation. that if the polynomial standard form of a quadratic equation has real coefficients and a real constant. How is this possible?
Answer 23-8
The coefficients and constant in the polynomial standard form of this equation are not all real numbers. They are not additive inverses. We can take the square root of both sides of the above equation. explain why not. we know that the ratio q /p is positive. and they are additive inverses There are two different complex roots. Its positive and negative square roots are therefore both real numbers.
Question 23-8
We have learned in the last several chapters (but not explicitly stated in full. Here is an example of such an equation in binomial factor form: (x + j 2)(x + j3) = 0 The roots of this equation are x = −j 2 or x = −j3.
Answer 23-7
A quadratic equation can have roots that are pure imaginary but not additive inverses.508 Review Questions and Answers
Because p and q are both positive. we can multiply the product of binomials out: (x + j 2)(x + j 3) = x 2 + j 3x + j 2x + (j 2j 3) = x 2 + j 5x − 6
. we found a quadratic equation that has two pure imaginary roots that are not additive inverses. then one of these things must be true: • • • • There are two different real roots There is a single real root with multiplicity 2 There are two different pure imaginary roots. as we can verify by plugging them in. If not.
Question 23-7
Is it possible for the roots of a quadratic equation to be pure imaginary but not additive inverses? If so. until now!). To see that. getting x = ±[−1(q /p)]1/2 = ±(−1)1/2 [±(q /p)1/2] = ±j[(q /p)1/2] The roots are therefore x = j[(q /p)1/2] or x = −j[(q /p)1/2]. and they are conjugates
In Answer 23-7.

and neither of them is pure imaginary. and the stand-alone constant is pure imaginary. To convert this to polynomial standard form.Part Three 509
In this case. If this sort of situation is impossible. If not. The coefficient of x 2. the coefficient of x is complex but not pure imaginary. are real. These are non-conjugate complex numbers. The complete polynomial quadratic equation is x 2 + j5x − 6 = 0
Question 23-9
Is it possible for one root of a quadratic equation to be pure real and the other pure imaginary? If so. is possible! Suppose the roots are 1 + j and 2 + j. taking extra precautions to be sure that we don’t mess up with the signs. explain why. too. we multiply the product of binomials out: (x + 2)(x + j3) = x 2 + j3x + 2x + j6 = x 2 + (2 + j3)x + j6 Here. this is possible. as we can verify by plugging them in. We can construct a binomial factor quadratic with these numbers as roots by subtracting the roots from x. provide an example of such an equation in polynomial standard form. Here is an example of such an equation in binomial factor form: (x + 2)(x + j3) = 0 The roots of this equation are x = −2 or x = −j3. The complete polynomial quadratic equation is x 2 + (2 + j3)x + j6 = 0
Question 23-10
Is it possible for a quadratic equation to have two nonconjugate complex roots. the coefficient of x is imaginary. The coefficient of x 2 is real. like this: [x − (1 + j )][x − (2 + j )] = 0 When we multiply the left side of this equation out to obtain a polynomial. as well as the stand-alone constant. explain why not.
Answer 23-10
This. we obtain [x − (1 + j)][x − (2 + j)] = [x + (−1) + (−j)][x + (−2) + (−j)] = x 2 + (−2x) + (−jx) + (−x) + 2 + j + (−jx) + j 2 + (−1) = x 2 + (−3x) + (−j 2x) + j 3 + 1 = x 2 + (−3 − j 2)x + (1 + j 3)
.
Answer 23-9
Yes. provide an example of such an equation in polynomial standard form. neither or which is pure imaginary? If so.

b. the coefficient of x is complex. The function has no real zeros if and only if the parabola doesn’t intersect the x axis at all.
Question 24-2
Suppose we see a quadratic function written as shown in Question 24-1. and c are real numbers with a ≠ 0: y = ax 2 + bx + c The graph of this function in Cartesian coordinates is always a parabola that opens either straight upward or straight downward. y) on the Cartesian plane by plugging in various values of x and calculating the results for y. Parabolas that open downward always have an absolute maximum. If the function has two real zeros where x = p and x = q. b. the vertex point) of the parabola? Let’s call it xv in this example. with specific numbers in place of a. and c. and is equal to the sum of the values divided by 2: xv = (p + q) / 2
. and a. That’s also known as the arithmetic mean. Imagine a quadratic function in which x is the independent variable and y is the dependent variable. How can we tell which way the parabola opens by simply looking at a specific function of this type?
Answer 24-1
The parabola opens straight upward if and only if a > 0. We plot several points (x.
Question 24-3
Parabolas that open upward always have an absolute minimum. y is the dependent variable. what is the x-value of the absolute maximum or minimum (that is. Its graph is a parabola. The parabola opens straight downward if and only if a < 0.
Answer 24-3
The value xv is the average of the two zeros.510 Review Questions and Answers
Here. assuming we plot enough points to get a “clear picture” of the parabola?
Answer 24-2
The quadratic function has two different real zeros if and only if the parabola crosses the x axis twice. The function has one real zero with multiplicity 2 if and only if the parabola is tangent to (“brushes up against”) the x axis at the absolute maximum point or the absolute minimum point. How can we determine how many real zeros the function has. the coefficient of x 2 is real. The complete polynomial quadratic is x 2 + (−3 − j 2)x + (1 + j 3) = 0
Chapter 24
Question 24-1
Consider the general form of a quadratic function where x is the independent variable. and the stand-alone constant is complex.

That’s also the x-value of the real zero. can be found by the formula xv = −b /2a
. what is xv. we have (x + 2)(x − 4) = 0 The zeros of the function are the same as the roots of this quadratic. To find the vertex point. we get y = x 2 −2x − 8 Because the coefficient of x 2 is positive. let’s remember the general polynomial standard form for a quadratic function: y = ax 2 + bx + c The x-coordinate of the vertex point.
Question 24-6
What are the real zeros of the function stated in Question 24-5? What are the coordinates (xv. we must get the right side of the equation in polynomial standard form by multiplying the binomials. we can see that these roots are x = −2 or x = 4.
Question 24-5
Suppose we come across the following quadratic function in binomial factor form. where x is the independent variable and y is the dependent variable: y = (x + 2)(x − 4) Does the parabola representing this function in Cartesian coordinates open upward or downward?
Answer 24-5
To determine this. Therefore.Part Three 511 Question 24-4
Imagine another quadratic function in which x is the independent variable and y is the dependent variable. When we do that. If we set it equal to 0. yv) of the vertex point in its graph? Is the vertex an absolute maximum or an absolute minimum?
Answer 24-6
The zeros can be seen by looking at the original form of the function. getting a quadratic equation in x. the parabola opens upward. the x-value of the vertex point on its graph?
Answer 24-4
When a quadratic function has only one real zero. If this function has a single real zero with multiplicity 2 where x = p. the parabola is tangent to the x axis at the vertex point. Without doing any algebra or arithmetic. xv. The right side of that equation is a product of binomials. xv = p.

Once again. a = 1 and b = −2. The vertex point is at (1. we get yv = (xv + 2)(xv − 4) = (1 + 2)(1 − 4) = 3 × (−3) = −9 Therefore. where x is the independent variable and y is the dependent variable: y = −3x 2 + 7x − 11 Does the parabola representing this function in Cartesian or rectangular coordinates open upward or downward? How many real zeros does the function have?
Answer 24-8
The parabola opens downward because the coefficient of x 2 is negative. −9).
Question 24-7
Based on the information in Answers 24-5 and 24-6. To determine how many real zeros the function has. This vertex is an absolute minimum because.
Question 24-8
Suppose we come across the following quadratic function in binomial factor form. the parabola opens upward. recall the general polynomial standard form for a quadratic function: y = ax 2 + bx + c Here. 30-1. yv) = (1. −9). we have a = −3.512 Review Questions and Answers
In this quadratic. Knowing all this. how can we sketch an approximate graph of the quadratic function stated in Question 24-5?
Answer 24-7
We have found that the zeros of the function are x = −2 and x = 4. b = 7. 0) and (4. 0). and c = −11. and the graph is a parabola that opens upward. as we found in Answer 24-5. we can sketch the graph as shown in Fig. we can evaluate the discriminant d. Therefore d = b2 − 4ac = 72 − 4 × (−3) × (−11) = 49 − 132 = −83
. The points representing them are at (−2. (xv. Therefore xv = −(−2) / (2 × 1) = 2/2 =1 Plugging this in and working out the arithmetic using the product of binomials.

The fact that d < 0 tells us that the function has no real zeros.0)
x –6 –4 –2 –4 2 6
–8 –10 (1.0) 2 (4.–9)
Figure 30-1 Illustration for Answer 24-7.
Question 24-9
What are the coordinates (xv. just as it tells us that the quadratic equation −3x 2 + 7x − 11 = 0 has no real roots. can be found by the formula xv = −b /2a
. This is the
graph of y = x2 −2x − 8. yv) of the vertex point in the graph of the function stated in Question 24-8? Is the vertex an absolute maximum or an absolute minimum?
Answer 24-9
The x-coordinate of the vertex point. xv.Part Three 513 y 6 4 (–2.

This is the
graph of y = −3x2 + 7x − 11.Part Three 515 1 2 x –4 –8 (0. the equation can be written like this: ax 3 + bx 2 + cx + d = 0 where a.92).–11) (2. On the x axis. c. On the y axis. 30-2. let’s use rectangular coordinates where each horizontal division is 1/4 unit and each vertical division is 2 units.–9)
–16 y
(7/6. b. and d are real numbers. The result is shown in Fig. −83/12) as (1. each increment is 1/4 unit.
Armed with this information. −6.
Chapter 25
Question 25-1
What is the polynomial standard form of a cubic equation in the variable x?
Answer 25-1
When a cubic equation is in polynomial standard form and the variable is x.–83/12)
Figure 30-2 Illustration for Answer 24-10.
Question 25-2
What is the binomial-cubed form of a cubic equation in the variable x ?
. We can use a calculator as a “point-plotting aid” and approximate the vertex (7/6. and a ≠ 0. To fit it into a neat space. each increment is 2 units. with the origin all the way up at the top. we can sketch the graph.17.

It is the solution to the equation we obtain when we set the binomial equal to 0: ax + b = 0 When we subtract b from both sides and then divide through by a (which is okay because we know that a ≠ 0).
Question 25-3
What does the binomial-cubed equation from Answer 25-2 look like when multiplied out into polynomial standard form?
Answer 25-3
Let’s begin by separating the left side of the equation into a product of three identical binomials: (ax + b)(ax + b)(ax + b) = 0 When we multiply the second two binomials together and then simplify the result into a trinomial. we get the equation (ax + b)(a2x 2 + 2abx + b2) = 0 Multiplying the binomial by the trinomial and simplifying gives us a3x3 + 3a2bx 2 + 3ab2x + b3 = 0
Question 25-4
How many real roots does the equation stated in Answer 25-2 have? What is that root. and b is a real number.516 Review Questions and Answers Answer 25-2
An equation in binomial-cubed form can always be written like this when x is the variable: (ax + b)3 = 0 where a is a nonzero real number. or what are they? What is the real solution set X?
Answer 25-4
There is one real root with multiplicity 3. we get x = −b /a The real solution set is therefore X = {−b /a}
.

Part Three 517 Question 25-5
What is the binomial-factor form of a cubic equation in the variable x?
Answer 25-5
An equation in binomial-factor form can always be written like this when x is the variable: (a1x + b1)(a2x + b2)(a3x + b3) = 0 where a1. b2. and b1. They are the solutions to the equations we obtain when we set the binomials equal to 0: a1x + b1 = 0 a2x + b2 = 0 a3x + b3 = 0
. That gives us a1a2a3x 3 + a1a2b3x 2 + a1b2a3x 2 + a1b2b3x + b1a2a3x 2 + b1a2b3x + b1b2a3x + b1b2b3 = 0 When we group the terms for x 2 and x together and apply the distributive law for multiplication over addition. a2. we obtain (a1x + b1)(a2a3x 2 + a2b3x + b2a3x + b2b3) = 0 Now let’s multiply the binomial by the polynomial on the left side. When we do that.
Question 25-6
What does the binomial-factor equation from Answer 25-5 look like when multiplied out into polynomial standard form?
Answer 25-6
Let’s start by multiplying the second and third binomials together. and a3 are nonzero real numbers. we get the polynomial standard form a1a2a3x 3 + (a1a2b3 + a1b2a3 + b1a2a3)x 2 + (a1b2b3 + b1a2b3 + b1b2a3)x + b1b2b3 = 0
Question 25-7
How many real roots does the equation stated in Answer 25-5 have? What is that root. and b3 are real numbers. or what are they? What is the real solution set X ?
Answer 25-7
There are three real roots.

−b2/a2. of the roots in Answer 25-7 to be the same? If so. or even all three.518 Review Questions and Answers
When we subtract the “b-constant” from both sides in each of these equations and then divide through by the “a-coefficient” (which is okay because we know that a1. Here’s an example: (x − 3)(x + 2)(−2x − 4) = 0 In this equation. and d are real numbers. b. The roots for each binomial. give examples. we get these roots: x = −b1/a1 or x = −b2/a2 or x = −b3/a3 The real solution set is therefore X = {−b1/a1. as before. are x = 5 or x = 5 or x = 5
Question 25-9
What is the binomial factor rule?
Answer 25-9
Imagine that we come across a cubic equation and we put it into the polynomial standard form. then the cubic equation has one real root with multiplicity 3.
.
Answer 25-8
Yes. explain why not. a2. In order from left to right. like this: ax3 + bx 2 + cx + d = 0 where a. c. the root for each binomial is the value of x that makes the binomial equal to 0. The binomial factor rule tells us that a real number k is a root of this equation if and only if (x − k) is a factor of the polynomial. one of them of multiplicity 2. Consider this: (x − 5)(2x − 10)(−3x − 15) = 0 Here. and a3 are all nonzero). those values are x = 3 or x −2 or x = −2 If all three of the roots are identical. If not. then the cubic equation has a total of two real roots. and a ≠ 0. this can happen. −b3/a3}
Question 25-8
It is possible for two. the root for each binomial is the x-value that makes it 0. in order from left to right. If two of the roots are identical.

the leading coefficient. a2.
Chapter 26
Question 26-1
What is the polynomial standard form of an nth-degree equation in the variable x. two real roots. There must always be at least one. In addition. must not be equal to 0. or three real roots.. can have one real root. we get the equation 0x n + an-1x n−1 + an-2x n−2 + ··· + a1x + b = 0 The term for xn has vanished. but there can never be more than three. a3.Part Three 519 Question 25-10
What is the smallest number of real roots that a single-variable cubic equation with real-number coefficients and a real-number constant can have? What is the largest number of real roots that such an equation can have?
Answer 25-10
A cubic equation in one variable. an. where n is a natural number larger than 3?
Answer 26-1
The polynomial standard form of such an equation is an x n + an-1x n−1 + an-2x n−2 + ··· + a1x + b = 0 where a1. what would happen if the coefficient an. leaving us with the polynomial standard form for a single-variable equation of degree n − 1: an-1x n−1 + an-2x n−2 + ··· + a1x + b = 0
Question 26-3
What is the binomial-to-the-nth form of an nth-degree equation in the variable x ?
.
Question 26-2
In an equation of the form shown in Answer 26-1. were equal to 0?
Answer 26-2
If an = 0.. . and b are real numbers. by which xn is multiplied. with real coefficients and a real constant. an.

(x 2 − 6x + 9)2 = 0
Answer 26-5
Let’s set the quantity (x 2 − 6x + 9) equal to 0.
Question 26-4
How many real roots does the equation stated in Answer 26-3 have? What is that root. It is the solution to the equation we obtain when we set the binomial equal to 0: ax + b = 0 When we subtract b from both sides and then divide through by a (which is okay because we know that a ≠ 0). the equation is of higher degree. the equation is cubic. If n > 3. the equation is of the first degree. or what are they? What is the real solution set X ?
Answer 26-4
There is one real root with multiplicity n. if n = 3. we get x = −b /a The real solution set is therefore X = {−b /a}
Question 26-5
Find all the real roots of the following equation. n can be any positive integer. the equation is quadratic. b is a real number. so it becomes the quadratic equation x 2 − 6x + 9 = 0 We can factor this into (x − 3)2 = 0
. state the multiplicity of each. Theoretically. if n = 2. and state the real solution set X. If n = 1.520 Review Questions and Answers Answer 26-3
An nth-degree equation in binomial-to-the-nth form can always be written like this when x is the variable: (ax + b)n = 0 where a is a nonzero real number. and n is a positive integer.

. bn are real stand-alone constants.. a2. or what are they? What is the real solution set X ?
Answer 26-7
There are n real roots.. an are nonzero real numbers.
Question 26-7
How many real roots does the equation stated in Answer 25-5 have? What is that root.. b2. and it has multiplicity 4. a2. we get these roots: x = −b1/a1 or x = −b2/a2 or x = −b3/a3 .. an are all nonzero).Part Three 521
If we substitute the quantity (x − 3)2 for the trinomial in the original equation.. a3. b3. . They are the solutions to the equations we obtain when we set the binomials equal to 0: a1x + b1 = 0 a2x + b2 = 0 a3x + b3 = 0 ↓ anx + bn = 0 When we subtract the “b-constant” from both sides in each of these equations and then divide through by the “a-coefficient” (which is okay because we know that a1. ...
Question 26-6
What is the binomial factor form of an nth-degree equation in the variable x?
Answer 26-6
Suppose that a1. or x = −bn/an
.. we get [(x − 3)2]2 = 0 which can be simplified to (x − 3)4 = 0 We can solve by setting the binomial equal to 0: x−3=0 This first-degree equation resolves to x = 3. Let x be the variable in the following equation: (a1x + b1)(a2x + b2)(a3x + b3) ··· (anx + bn) = 0 This is the binomial factor form for an nth-degree equation in the variable x. This is the only real root of the original higherdegree equation. . The real solution set is X = {3}... and b1. a3.

. ··· −bn/an}
Question 26-8
What are the real roots of the following equation? What is the multiplicity of each root? What is the real solution set X? What is the degree of the equation? (x + 4)(2x − 8)2(3x)5 = 0
Answer 26-8
We take each binomial individually. set it equal to 0. a2.
Question 26-9
What is the largest number of real roots that a single-variable equation of the nth degree can have? What is the largest number of real or complex roots that such an equation can have?
Answer 26-9
A single-variable equation of the nth degree can have. The root 4 has multiplicity 2. in this case 1 + 2 + 5. x = 4. . an. The root −4 has multiplicity 1. −b2/a2. and then solve the resulting first-degree equations: x+4=0 2x − 8 = 0 3x = 0 These equations resolve to x = −4. at most. or 8. n roots in total. The root 0 has multiplicity 5. and b are nonzero rationals. What is that process?
.522 Review Questions and Answers
The real solution set is therefore X = {−b1/a1. −b3/a3. a3. 4. and x = 0 respectively. 0}. The degree of the equation is the sum of the exponents attached to the factors. Therefore. considering the real-number roots and the complex-number roots combined.
Question 26-10
There’s a way to find all the rational-number roots of an nth-degree equation in the single variable x when it appears in the polynomial standard form anx n + an−1x n−1 + an−2x n−2 + ··· + a1x + b = 0 where a1. and n is a positive integer greater than 3.. the real roots of the original equation are x = −4 or x = 4 or x = 0 and the real solution set is X = {−4.

• If none of the numbers r produces a remainder of 0. and so on. in the order shown below. Call those ratios r. and then the quadratic formula can be used to find its roots. Each of these binomials is a factor of the original equation.
Chapter 27
Question 27-1
Suppose we’re confronted with a pair of equations in two variables. a3. The process can be tedious. but it’s often more likely to produce useful results than tackling the equation “head-on” or hoping for an intuitive breakthrough. multiply the equation through by the smallest constant that will turn them all into integers. • List all of the rational roots. That factor can be set equal to 0. • Find all the positive and negative integer factors of an. • Create binomials of the form (x − r1). and one or both of the equations is nonlinear. How can we solve these equations as a two-by-two system?
Answer 27-1
When we want to solve a general two-by-two system of equations. we can be sure that none of the roots associated with it are rational. • Write down all the possible ratios m /n. Call them r1. (x − r2). • Find the root(s) of that equation. Neither of those roots will be rational. • Mix the independent-variable parts of the equations to get an equation in one variable. • Decide which variable to call independent. then every one of those numbers is a rational root of the equation. • Morph both equations so they express the dependent variable in terms of the independent variable. then the original equation has no rational roots. we’ll get one or more binomial factors and a quadratic factor. and so on.Part Three 523 Answer 26-10
We must do the following things. a2. or an equation in binomial factor form. we’ll get one or more binomial factors and a cubic or higher-order polynomial factor. Call those factors m.. an. to see if we get a remainder of 0. . If they aren’t. • Find all the positive and negative integer factors of b. we can go through these steps in order. • With synthetic division.. check every r. They might even be complex. • If we’re lucky. • If we’re unlucky. this process will give us an equation in binomial to the nth form. If we set the polynomial factor equal to 0. • Make sure that a1. r3. • If we’re less lucky. and which one to call dependent. and b are all integers. r2. • If one or more of the ratios r produces a remainder of 0. one at a time. Call those factors n. (x − r3).
.

• Express the solution(s) as one or more ordered pairs. x = 0. When we plug this into either of the original equations. When we mix the right sides.0). let x be the independent variable.
Question 27-3
Consider the following pair of quadratic equations as a two-by-two system: y = x2 − 1 and y = −x 2 + 1 How can we find the real solutions of this system?
Answer 27-3
Again.
Question 27-2
Consider the following pair of quadratic equations as a two-by-two system: y = x2 and y = −x 2 How can we find the real solutions of this system?
Answer 27-2
Let’s call x the independent variable. and calculate the corresponding value(s) of the dependent variable. Therefore.524 Review Questions and Answers
• Plug the root(s) into one of the morphed original equations. we get y = 0. the single real solution to this system is (0. we get x 2 − 1 = −x 2 + 1
. Then we can mix the right sides of the equations to get x 2 = −x 2 Adding x 2 to each side. we get 2x 2 = 0 Dividing through by 2 gives us x2 = 0 This equation has one real root.

we get x 2 = (b2 − b1) / (a1 − a2) This means that x = [(b2 − b1) / (a1 − a2)]1/2 or x = −[(b2 − b1) / (a1 − a2)]1/2
. which we know is okay because we’ve been told that a1 ≠ a2. Plugging x = 1 into the first original equation. we obtain x2 = 1 It’s apparent. without doing any algebra. let’s call x the independent variable. we get a1x 2 + b1 = a2x 2 + b2 We can subtract a2x 2 from each side and then apply the distributive law to get (a1 − a2)x 2 + b1 = b2 Subtracting b1 from each side produces (a1 − a2)x 2 = b2 − b1 Dividing through by the quantity (a1 − a2).Part Three 525
Adding the quantity (x 2 + 1) to each side.0) and (−1. we get 2x 2 = 2 Dividing this equation through by 2. How can we find the real solutions of this system?
Answer 27-4
Once again.
Question 27-4
Consider the following pair of quadratic equations as a two-by-two system: y = a1x 2 + b1 and y = a2x 2 + b2 where a1 and a2 are nonzero real numbers that are not equal to each other. that the roots of this are x = 1 or x = −1. This system therefore has two real solutions. (1. we again get y = 0.0). When we mix the right sides. we get y = 0. Plugging x = −1 into that same equation. and b1 and b2 are real numbers.

we get y = a2(b2 − b1) / (a1 − a2) + b2 Multiplying the right-hand side out. we obtain the same result as before: y = (a1b2 − a2b1) / a1a2
Question 27-6
Consider the following pair of quadratic equations as a two-by-two system: y = (x + 1)2 and y = (x − 1)2 How can we find the real solutions of this system?
. we get y = (a1b2 − a1b1) / (a1 − a2) + b1 Expressing b1 as the ratio b1/1. we obtain y = (a1b2 − a2b1) / a1a2 Writing these solutions as ordered pairs is tricky and messy. Then we can write x = ±[(b2 − b1) / (a1 − a2)]1/2 and y = (a1b2 − a2b1) / a1a2
Question 27-5
What happens when we plug the solution for x 2 into the second original equation in the above system (instead of the first one. so we might as well substitute directly for x 2 into the first original equation to obtain y = a1(b2 − b1) / (a1 − a2) + b1 Multiplying the right-hand side out. which we already did) and solve for y?
Answer 27-5
When we substitute directly for x 2 into the second original equation. We’ll get the same thing in both cases. applying the general rule for adding ratios. We might as well state the x and y values separately and take advantage of the plus-or-minus sign. we get y = (a2b2 − a2b1) / (a1 − a2) + b2 Expressing b2 as the ratio b2/1. and then canceling out identical terms that subtract from each other.526 Review Questions and Answers
When we plug either of these into the first original equation. and then canceling out identical terms that subtract from each other. we have to square it. applying the general rule for adding ratios.

we obtain y = a 2x 2 + 2abx + b 2 and y = a 2x 2 − 2abx + b 2 Setting the right sides equal gives us the “mixed quadratic” a 2x 2 + 2abx + b 2 = a 2x 2 − 2abx + b 2
. we get y = 1. When we do that. Therefore. How can we find the real solutions of this system?
Answer 27-7
Again.” When we substitute this value for x into either of the original equations. the system has the single real solution (0. neither of which are equal to 0. When we do that.1). let’s multiply out the right sides of the binomial-squared equations. obtaining the quadratics in polynomial standard form. we obtain 2x = −2x We can add 2x to each side and then divide through by 4 to get x = 0 as the sole root of this “mixed quadratic. we get y = x 2 + 2x + 1 and y = x 2 − 2x + 1 Mixing the right sides of these quadratics gives us x 2 + 2x + 1 = x 2 − 2x + 1 Subtracting the quantity (x 2 + 1) from each side.Part Three 527 Answer 27-6
The best approach here is to multiply out the right sides of both of the binomial-squared equations.
Question 27-7
Consider the following pair of quadratic equations as a two-by-two system: y = (ax + b)2 and y = (ax − b)2 Where a and b are real numbers.

” We can substitute this value for x into the original equations. and we mix them to get a single equation in one variable. and another solution with multiplicity 2.
Question 27-9
Consider the following pair of equations: y = (x + 1)3 and y = x 3 + 2x 2 + x How can we find the real solutions to this two-by-two system?
Answer 27-9
Let’s multiply out the first equation to get it into polynomial standard form. so we can be sure that 4ab ≠ 0). When we solve that “mixed” equation. and the other of which has multiplicity 2. the system becomes y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 2x 2 + x We can mix the right sides of these equations. Therefore. we get two different roots. What does this say about the solutions of the original system?
Answer 27-8
The solutions of the original two-by-two system have the same multiplicity pattern as the roots of the “mixed” equation. In this case. the system has the single solution (0. getting x 3 + 3x 2 + 3x + 1 = x 3 + 2x 2 + x
.
Question 27-8
Imagine that we’re trying to solve a general two-by-two system of equations.b2). one of which has multiplicity 1.528 Review Questions and Answers
When we subtract the quantity (a 2x 2 + b 2) from each side. obtaining y = b2 in both cases. When we do that. When we do that. we obtain x = 0 as the only root of the “mixed quadratic. that means there is one solution with multiplicity 1. we get 2abx = −2abx We can add 2abx to each side and then divide through by 4ab (which is “legal” because we’ve been told that a ≠ 0 and b ≠ 0.

0). we obtain 3x 2 + 9x + 7 = 0
. we get x 3 + 3x 2 + 3x + 1 = x 3 + 6x 2 + 12x + 8 Subtracting the entire left side from the right side and then switching right-to-left. the original system has the single real solution (−1. When we do that. with multiplicity 2. we obtain x 2 + 2x + 1 = 0 which factors into (x + 1)2 = 0 This equation has the single real root x = −1. if any exist?
Answer 27-10
Let’s multiply both of these equations out to get cubics in polynomial standard form. we get y = (x + 1)3 = (−1 + 1)3 = 03 =0 Therefore. with multiplicity 2. That gives us y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 6x 2 + 12x + 8 When we mix the right sides of these equations. When we plug that into the first original equation.Part Three 529
Now let’s subtract the entire right side of this equation from the left side.
Question 27-10
Consider the following pair of equations: y = (x + 1)3 and y = (x + 2)3 How can we find the real solutions of this two-by-two system.

Table 30-1 compares some values of x.0).
Chapter 28
Question 28-1
How do we graph a general two-by-two system of equations when we want to see approximately where the curves intersect at the real solutions. and some values of
Table 30-1. Then we plot the real solution point or points. we can calculate several ordered pairs for both functions individually. After that. if any exist. It can be helpful to put the values in a table. we plot the rest of the points based on the values in the table we’ve created.
x2 4 1 0 1 4 −x 2 −4 −1 0 −1 −4
x −2 −1 0 1 2
.
Question 28-2
Consider the system of equations we solved in Answer 27-2: y = x2 and y = −x 2 How can we sketch an approximate graph of this system.
Selected values for graphing the functions y = x 2 and y = −x 2. The x-values of any solutions we can derive for the original system will not be real numbers. if any exist. including the real solutions. (0. no real solutions exist. The bold entry indicates the real solution. but we don’t need a lot of precision?
Answer 28-1
First. on the coordinate grid. some values of the first function. showing the real solution?
Answer 28-2
We can tabulate and plot several points in both functions including the real solution. telling us that it has no real roots. Finally. we fill in the lines or curves for both functions. we figure out the scales we should have on our graph so as to provide a good “picture” of the situation. Next.530 Review Questions and Answers
The discriminant of this quadratic is negative. Therefore.

each increment represents 1 unit. Table 30-2 compares some values of x.
Question 28-3
Consider the system of equations we solved in Answer 27-3: y = x2 − 1 and y = −x 2 + 1 How can we sketch an approximate graph of this system.0). Figure 30-3 shows the curves and the solution point. (1. The real-number solution appears as a point where the curves intersect.Part Three 531
y
(0. the second function is graphed as a dashed curve.0) and (−1.
. Figure 30-4 shows the curves and the solution points. On both axes. and some values of the second function. The first function
is graphed as a solid curve.0)
x
Figure 30-3 Illustration for Answer 28-2. each increment is 1 unit.
the second function. On both axes. each increment represents 1 unit. showing the two real solutions?
Answer 28-3
We can tabulate and plot several points in both functions including the real solutions. On both axes. some values of the first function.

Real-number solutions appear as points where the curves intersect. and the positive x-value
y
(–1. But in the second case.
x2 − 1 3 0 −1 0 3 −x 2 + 1 −3 0 1 0 −3
x −2 −1 0 1 2 Question 28-4
When we compare the systems stated in Questions 28-2 and 28-3 and graphed in Figs. What will happen if we move the upward-opening parabola. the intersection points will move farther from each other. The first function
is graphed as a solid curve. further up by the same distance? How will the equations in the system change if we do this?
Answer 28-4
If we move the parabolas this way. 30-3 and 30-4. 30-3) to “break in two” (Fig. we can see that the curves have the same shapes in both situations. The negative x-value of one real solution will become more negative. and move the downward-opening parabola.0)
x
Figure 30-4 Illustration for Answer 28-3.532 Review Questions and Answers
Table 30-2. shown by the dashed curve.
. each increment is 1 unit. This has caused the single intersection point (Fig. 30-4). On both axes. further down.0)
(1. while the downward-opening parabola has been moved vertically up by 1 unit. Bold entries indicate real solutions. the upward-opening parabola has been moved vertically down by 1 unit. the second function is graphed as a dashed curve. shown by the solid curve.
Selected values for graphing the functions y = x 2 − 1 and y = −x 2 + 1.

each increment represents 1 unit. Suppose that we move the upward-opening parabola even further straight down. On both axes. On both axes. The negative constant in the first equation will become more negative. The y-values of both solutions will remain at 0. Real-number solutions appear as points where the curves intersect. 30-5. and the positive constant in the second equation will stay the same. the second function is graphed as a dashed curve. The negative constant in the first equation will become more negative. The stand-alone constants in the equations will change. On both axes.
of the other real solution will become more positive to the same extent.
Question 28-5
Let’s modify the system presented in Question 28-4 and graphed in Fig. The solution points will move off the x axis into the third and fourth quadrants of the coordinate plane. each increment represents 1 unit.
Question 28-6
Consider the system of equations we solved in Answer 27-6: y = (x + 1)2
. Figure 30-5 shows an example. and the positive x-value of the other real solution will become more positive to the same extent. This assumes that we move the upward-opening parabola exactly in the negative-y direction. and the positive constant in the second equation will become more positive to the same extent. The first function
is graphed as a solid curve. but leave the downwardopening parabola in the same place. What will happen to the solution points? How will the equations in the system change?
Answer 28-5
The intersection points will move even farther from each other. The negative x-value of one real solution will become more negative. The y-values of both solutions will become negative to an equal extent.Part Three 533
y
Solution
Solution
x
Figure 30-5 Illustration for Answer 28-4. each increment is 1 unit. the points will stay on the x axis. Figure 30-6 shows an example.

On the y axis. Figure 30-7 shows the curves and the solution point.
Selected values for graphing the functions y = (x + 1)2 and y = (x − 1)2. the second function is graphed as a dashed curve.
and y = (x − 1)2 How can we sketch an approximate graph of this system. Real-number solutions appear as points where the curves intersect. (0.
Table 30-3. On the x axis. The bold entry indicates the real solution.1). Table 30-3 compares some values of x. some values of the first function.
(x + 1)2 4 1 0 1 4 9 16 (x − 1)2 16 9 4 1 0 1 4
x −3 −2 −1 0 1 2 3
. On both axes. each increment is 1/2 unit. each increment is 2 units. and some values of the second function. each increment is 1 unit.534
Review Questions and Answers
y
x
Solution
Solution
Figure 30-6 Illustration for Answer 28-5. showing the real solution?
Answer 28-6
We can tabulate and plot several points in both functions including the real solution. The first function
is graphed as a solid curve.

Part Three 535
y
(0. On the y axis. producing this two-by-two system: y = (x + 1)2 and y = −(x − 1)2 How will this affect the graph of the second equation. Suppose we subtract 12 from the right side of the second equation. The first function
is graphed as a solid curve.
Question 28-8
Let’s modify the system of equations stated in Question 28-6 in a different way. we multiply all values of the function by −1. Suppose we multiply the right side of the second equation by −1. each increment is 1/2 unit. producing this two-by-two system: y = (x + 1)2
. each increment is 2 units. shown by the dashed curve? How will it affect the real solution set?
Answer 28-7
If we multiply the right side of this equation by −1. On the y axis. each increment is 1/2 unit. Figure 30-8 shows the result. We can see that the system has no real solutions because the curves don’t intersect. Question 28-7
Let’s modify the system of equations stated in Question 28-6.1)
x
Figure 30-7 Illustration for Answer 28-6. This inverts the entire graph of the function with respect to the x axis. The real solution set is empty. On the x axis. each increment is 2 units. On the x axis. the second function is graphed as a dashed curve. The real-number solution appears as a point where the curves intersect.

showing the real solution?
. but it has changed. Figure 30-9 shows the result. This moves the entire graph of the function vertically down by 12 units. shown by the dashed curve? How will it affect the real solution set?
Answer 28-8
If we subtract 12 from the right side of this equation.
y
No intersection points
x
No real solutions
and y = (x − 1)2 − 12 How will this affect the graph of the second equation.536 Review Questions and Answers
Figure 30-8 Illustration for Answer
28-7. The curves do not intersect. the second function is graphed as a dashed curve. indicating that the system has no real solutions. each increment is 1/2 unit. On the y axis. (For extra credit. each increment is 2 units. If we want to find the solution. you can do this. This graph suggests that the resulting system still has one real solution. On the x axis. we must solve the new system. starting all over again from scratch. The first function is graphed as a solid curve. we reduce all values of the function by 12. On the x axis. On the y axis. each increment is 2 units. each increment is 1/2 unit.)
Question 28-9
Consider the system of equations we solved in Answer 27-9: y = (x + 1)3 and y = x 3 + 2x 2 + x How can we sketch an approximate graph of this system.

each increment is 1/2 unit. we must solve the new system algebraically.
(x + 1)3 −8 −1 0 1 8 27 x 3 + 2x 2 + x −12 −2 0 0 4 18
x −3 −2 −1 0 1 2
. (−1. On the y axis. some values of the first function.Part Three 537
y
Solution
x
Figure 30-9 Illustration for Answer 28-8. and some values of the second function.0). each increment is 2 units. the second function is graphed as a dashed curve. Figure 30-10 shows the curves and the solution point. To find its coordinates.
Table 30-4.
Selected values for graphing the functions y = (x + 1)3 and y = x 3 + 2x 2 + x. each increment is 5 units. On the x axis. The solution point has moved. On the y axis. The first function
is graphed as a solid curve. On the x axis. The bold entry indicates the real solution.
Answer 28-9
We can tabulate and plot several points in both functions including the real solution. Table 30-4 compares some values of x. each increment is 1/2 unit.

where we found that there are no real solutions: y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 6x 2 + 12x + 8 How can we sketch an approximate graph of this system. On the x axis. showing that there are no real solutions?
Answer 28-10
We can tabulate and plot enough points in both functions to get clear images of their curves. each increment is 1 unit. and to show that they do not intersect (although they come close).
Chapter 29
Question 29-1
If we say that the common logarithm of a certain number p is equal to q. On the y axis.0)
x
Figure 30-10 Illustration for Answer 28-9. each increment is 10 units. the second function is graphed as a dashed curve. The real-number solution appears as a point where the curves intersect. Figure 30-11 shows the curves. Question 28-10
Consider the system of equations we evaluated in Answer 27-10. Table 30-5 compares some values of x. some values of the first function. If we say that the common log of p is equal to q. and some values of the second function. The first function
is graphed as a solid curve. On the x axis.538 Review Questions and Answers
y
(–1. each increment is 5 units. we mean p = 10q
. On the y axis. each increment is 1/2 unit. what do we mean?
Answer 29-1
The common logarithm (or common log) of a number is the power to which we must raise 10 to get that number.

what is the common log of 10? Of 100? Of 1.
x −5 −4 −3 −2 −1 0 1 2 x 3 + 3x 2 + 3x + 1 −64 −27 −8 −1 0 1 8 27 x 3 + 6x 2 + 12x + 8 −27 −8 −1 0 1 8 27 64
The common log is sometimes called the base-10 log. as that number grows larger and larger indefinitely?
Answer 29-2
The common log is related to a growing positive number like this: • The common log of 10 is 1.
y
No intersection points
x
No real solutions
. The curves do not intersect (although they come close!). Selected values for graphing the functions y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 6x 2 + 12x + 8.
Question 29-2
According to the definition in Answer 29-1.000? Of 10. • The common log of 100 is 2. so there are no real solutions.Part Three 539
Table 30-5. Figure 30-11 Illustration for Answer 28-10. the second function is graphed as a dashed curve.000? What happens to the common log of a positive number. because 101 = 10. because 10 is the base that we raise to various powers. On the y axis. each increment is 1 unit. because 102 = 100. This system has no real solutions. On the x axis. each increment is 10 units.
The first function is graphed as a solid curve.

The value of e is approximately 2.000 is −3. (Perhaps it’s non-real but complex. because 10−1 = 1/10. as the number approaches 0 from the positive direction? What happens to the common log of a shrinking positive real number when it actually becomes 0?
Answer 29-3
As the absolute value of a positive number keeps shrinking. The common log of 1/1. because 10−2 = 1/100. or maybe it’s some other kind of number entirely. because 104 = 10. As the shrinking positive number “closes in” on 0.000. its common log changes like this: • • • • The common log of 1 is 0. What is the common log of 1? Of 1/10? Of 1/100? Of 1/1000? What happens to the common log of a positive real number whose absolute value keeps shrinking. As a number gets larger without limit. it increases negatively—without limit.
Question 29-5
According to the definition in Answer 29-4.000 is 3. what is the natural log of e ? Of e 2? Of e 3? Of e4? What happens to the natural log of a number as that number grows larger without limit?
. e. because 10−3 = 1/1.
Question 29-3
According to the definition in Answer 29-1. • The common log of 10. because 100 = 1. so does its common log. its common log is no longer defined in the set of real numbers. because 103 = 1. however. The common log of 1/100 is −2. we mean p = eq The common log is sometimes called the base-e log.71828.000.540 Review Questions and Answers
• The common log of 1. what do we mean?
Answer 29-4
The natural logarithm (or natural log) of a number is the power to which we must raise Euler’s constant.000. The common log of 1/10 is −1. because e is the base that we raise to various powers. to get that number. that is. The size of the logarithm grows much more slowly than the size of the number.
As a shrinking positive number approaches 0. its common log becomes more negative. Evaluating it is beyond the scope of this book. so it cannot be fully written out in decimal form. It’s an irrational number. the common log decreases—that is. When the shrinking positive number actually reaches 0.000 is 4.)
Question 29-4
If we say that the natural logarithm of a certain number p is equal to q. If we say that the natural log of p is equal to q.

its natural log is no longer defined in the set of real numbers. These rules work for common logs as well as for natural logs. There is no limit to how large negatively the log can get. so does its natural log. because e 0 = 1. as long as we don’t change the base during the calculation!
Question 29-8
What is the common exponential of a number? What is the natural exponential of a number?
. The natural log of 1/e 2 is −2. or ratios into differences? Do these properties of logs depend on the base?
Answer 29-7
The logarithm of the product of two numbers is equal to the sum of their logarithms. because e 1 = e.)
Question 29-7
How can logarithms be used to change products into sums.
Question 29-6
According to the definition in Answer 29-4. The natural log of e 2 is 2. The natural log of e 3 3.
As a positive number approaches 0. When the shrinking positive number actually reaches 0. The natural log of e 4 is 4. (Perhaps it’s non-real but complex. Evaluating it is beyond the scope of this book. The logarithm of the ratio of two numbers is equal to the difference between their logarithms.
As a number gets larger without limit.Part Three 541 Answer 29-5
The natural log is related to a growing number like this: • • • • The natural log of e is 1. they work no matter what the base happens to be. The natural log of 1/e 3 is −3. In fact. or maybe it’s some other kind of number entirely. What is the natural log of 1? Of 1/e ? Of 1/e 2? Of 1/e 3? What happens to the natural log of a positive real number whose absolute value keeps shrinking? What happens to the natural log of a shrinking positive real number when it actually becomes 0?
Answer 29-6
As the absolute value of a positive number keeps shrinking. but the size of the log grows much more slowly than the size of the number. its natural log changes like this: • • • • The natural log of 1 is 0. The natural log of 1/e is −1. its natural log becomes more negative.

000 A calculator tells us that log10 100. But if we want to go through the motions of solving the above equation formally.000?
Answer 29-9
To find the number x whose common exponential is 100. When working with natural exponentials. we must find the power of 10 that gives us 100. We want to solve the equation 10 x = 100. Finding the number y whose natural exponential is 100. We want to find the power of e that gives us 100.000 The common log function “undoes” the common exponential function. we can take the common log (symbolized log10) of both sides.000 is a little more involved. obtaining log10(10 x) = log10 100.513. we get ln (e y) = ln 100. so we must solve the equation e y = 100. so we have y = ln 100.000 is exactly equal to 5.000 The natural log function “undoes” the natural exponential function.
Question 29-9
How can we find the number x whose common exponential is 100.000.000 A calculator tells us that y = 11.000.542 Review Questions and Answers Answer 29-8
The common exponential of a number is what we get when we raise 10 to a power equal to that number.000 If we take the natural log (symbolized ln) of both sides of this equation.
Question 29-10
How can we find the number x whose natural exponential is 1/e 5? How can we find the number y whose common exponential is 1/e 5?
. rounded off to three decimal places.000. The natural exponential of a number is what we get when we raise e to a power equal to that number. so we can simplify this equation to x = log10 100.000 It’s easy see that x = 5 in this case.000? How can we find the number y whose natural exponential is 100. it’s customary to call e the exponential constant.

171
. because 1/e 5 is just another way of writing e −5.006737947) = −2. we can take the natural log of both sides of the above equation. We want to find the power of 10 that gives us e −5. Rounding off the end result to three decimal places. despite the simplicity of this. which is the common log of 0. so we must solve the equation 10y = e −5 If we take the common log of both sides of this equation. but not much. we obtain log10 (10 y) = log10 (e −5) The common log function “undoes” the common exponential function. we must find the power of e that gives us 1/e 5. Now we have e x = e −5 Obviously.Part Three 543 Answer 29-10
To find the number x whose natural exponential is 1/e 5.006737947. We want to solve the equation e x = 1/e 5 This is almost trivial. rounded off to nine decimal places. obtaining ln (e x) = ln (e −5) We can simplify both sides to get x = −5 Finding the number y whose common exponential is 1/e 5 requires more work. this means x = −5. so we have y = log10 (e −5) A calculator tells us that e −5 = 0. we get y = log10 (0. we insist on solving formally and including every step. That ought to be plenty of digits to give us a good idea of the final answer.006737947. If.

(c) π (d) 641/2 (e) (64/9)1/2 5. (d) There are infinitely many solutions.25252525 . All of the following equations except one are generally true. (c) the equation has one imaginary root with multiplicity 2. If the discriminant in a quadratic equation is equal to 0. Which of the following is not a rational number? (a) 22/7 (b) 25. Which one is the exception? (a) (m + n)/p = m /p + n /p (b) p /(m − n) = p /m − p /n (c) p(m + n) = pm + pn (d) (m + n)p = mp + np (e) (m − n)p = mp − np 6. (c) (x. it means that (a) the equation has two different real roots. complex roots. (d) three solutions. (b) the equation has one real root with multiplicity 2. (e) the equation has no roots at all. (e) There is no solution. y) = (−4/3. −1/2). (e) infinitely many solutions.546 Final Exam
What is the solution to this system? (a) (x. y) = (0. 3. (b) one solution. (c) two solutions.. 2). (b) (x. 4. We are sure we haven’t made any mistakes along the way.
. y) = (3/4. 4). We come up with this matrix:
2 3 −1 2 3 −1 2 3 −1 8 −7 5
We can conclude that the original three-by-three system has (a) no solutions. and p is a nonzero integer.. Suppose that m and n are integers. (d) the equation has two different nonreal. Suppose we are trying to solve a three-by-three linear system using matrices.

) (a) Eight days. Fill in the blank to make the following statement true. In Fig. 12. FE-1. When we want to add two complex numbers. How many 24-hour days will pass between this moment and 12:00 noon on the fourth day of July. (e) None of the above. (e) Twelve days.Final Exam
547
7. getting a real number. getting a real number. 10. and sets A and D are disjoint. Suppose it’s 12:00 noon on the twenty-fifth day of June. getting a complex number. In Fig. with n ≠ 0. we must (a) add the real parts and multiply the imaginary parts. and sets A and D are disjoint. FE-1. C. (e) find their absolute values and add them. If we raise a to the mth power and then take the nth root of that quantity. (d) Sets A and C are disjoint. (d) add the real parts and the imaginary parts separately. getting a complex number. (b) 4-1/3. assuming A. (b) Nine days. and sets C and D are disjoint. (c) multiply the real parts and the imaginary parts separately. Also suppose that m and n are rational numbers. (b) Sets A and C are disjoint. which of the following statements is true. “Suppose that a is a nonzero number. C. B. assuming A.” (a) raise a to the power of (m + n) (b) raise a to the power of (m − n) (c) raise a to the power of mn (d) raise a to the power of m /n (e) raise a to the power of 1/(mn) 8. and sets B and D are disjoint. and D are all non-empty sets? (a) Sets A and B are disjoint. The numerical value of 13/(−3) is the same as the value of (a) −4-1/3. which of the following statements is true. in the same year and in the same time zone? (June has 30 days. (c) Ten days. (b) multiply the real parts and add the imaginary parts. (c) Sets B and C are disjoint. (d) Eleven days.
. we get the same result as if we ________. 9. B. getting a real number. and D are all non-empty sets? (a) B ⊆ A (b) B ∈ A (c) A ∩ B = A (d) A ∩ B = ∅ (e) A ∪ B = B 11.

(b) is endless and repeating. (d) for all possible values of x and y.Final Exam
549
How many real roots does this equation have? (a) None. 18. (b) 1/5. (b) One. as long as y ≠ 1. (b) for all possible values of x and y. (e) Infinitely many. (e) None of the above. it’s the same thing as raising that number to the power of (a) −5. (b) j. When we take the 5th root of a number. The decimal expansion of an irrational number between 0 and 1 (a) terminates after a finite number of digits. (d) can be converted to a ratio of two integers.
. (d) 25. as long as y ≠ 0. Which of the following is a 4th root of 1? (a) 1 (b) −1 (c) j (d) −j (e) All of the above 16. (d) Three. (c) −1/5. 19. Suppose x and y are real numbers. as long as x ≠ 0. as long as y ≠ −1. 17. (c) for all possible values of x and y. The quantity [x /(y + 1)]2 makes sense (a) for all possible values of x and y. (c) Two. 15. (e) for all possible values of x and y. (e) −25. (c) is endless and non-repeating. Consider the following equation: (x − j5)7 = 0 This equation has an imaginary root j5 with multiplicity (a) 1.

In Fig. what is being done to the original number 16n? (a) It is being repeatedly multiplied by 2. (d) j5.
Figure FE-2 Illustration for Final Exam
Questions 21 and 22. (e) 7. FE-2.
Positive numbers Start at 16n
Finish at n
4n
“Number reflector”
–2n
a
Negative numbers
.550 Final Exam
(c) 5. 20. What’s the binary equivalent of the decimal 127? (a) 1001000 (b) 1111111 (c) 1000001 (d) 10000000 (e) 11100111 21. (b) It is being repeatedly multiplied by −2.

it’s the same thing as (a) taking the 5th root of the quantity and then taking the reciprocal of the result. (d) All natural numbers are integers. (b) Two. (b) −6n. (d) It is being repeatedly divided by −2. Consider the following third-degree equation: (x − 1)(x + 7)2 = 0 How many real roots does this equation have? (a) None. where all the coefficients and constants are real numbers. (e) None of the above. When we take a quantity to the power of −1/5.
. 25. (c) Three. We want to solve these functions as a twoby-two system. (d) taking the −5th power of the quantity and then taking the reciprocal of the result. (c) −8n. the value of a is (a) impossible to determine without more information. In Fig. (e) It is being repeatedly squared. (b) One. (e) We can’t answer this without more information. (d) Infinitely many. 22. (c) Two. (c) All negative rational numbers are real. (b) taking the 5th power of the quantity and then taking the reciprocal of the result. Which of the following statements is false? (a) All integers are rational. What is the maximum number of real solutions that such a system can have? (a) One. 26. 24. (b) All real numbers are irrational. Imagine that we come across a cubic function and a linear function. (e) −12n. (d) −10n. 23. (e) All integers are real. (c) taking the −5th root of the quantity and then taking the reciprocal of the result.Final Exam
551
(c) It is being repeatedly divided by 2. FE-2.

and three. Imagine a two-by-two linear system in variables x and y. and infinitely many. a2. (d) One. and three. b2. and neither a1 nor a2 are equal to 0. Which of these integers is the smallest? (a) −8 (b) −2 (c) 0 (d) 3 (e) 7 28. and one.
. b1. Consider the following pair of equations as a two-by-two system: y = a1x + b1 and y = a2x 2 + b2x + c where a1. What are the smallest and largest numbers of elements that the solution set of such a system can have? (a) None. (c) solutions corresponding to the x-intercept of one line and the y-intercept of the other line. but not coincident. 27. where y is the dependent variable. (e) None. and two. (b) solutions corresponding to the x-intercepts of the lines. Such a system has (a) solutions corresponding to the y-intercepts of the lines. (d) no solutions. (b) None. 30. Suppose the graphs of the equations are parallel. lines in Cartesian coordinates. (e) Infinitely many. Which of these integers has the largest absolute value? (a) −8 (b) −2 (c) 0 (d) 3 (e) 7 29. (c) None.552 Final Exam
(d) Three. and c are real numbers. (e) infinitely many solutions.

. (e) The set of all irrational numbers. 35. We subtract −a from it. Suppose we have a positive integer a. 34. we create a singlevariable equation by mixing the independent-variable parts of the functions. (d) equal to −a − a. (c) equal to a + a. As we solve these functions as a two-by-two system. (e) There is one real solution with multiplicity 6. Suppose we see two cubic functions. (c) The set of all integers.” (a) raise a to the power of (m + n) (b) raise a to the power of (m − n) (c) raise a to the power of mn (d) raise a to the power of m /n (e) raise a to the power of 1/(mn) 33. The intersection of the null set with any other set is always equal to (a) that other set. what can we say about the multiplicities of the real solutions of the original system? (a) Nothing.Final Exam
553
31. (b) equal to a. different real solution with multiplicity 2. we discover that it can be written in binomial-cubed form. Based on that knowledge. (d) The set of all rational numbers. Which of the following sets is nondenumerable? (a) The set of all natural numbers. (d) There is one real solution with multiplicity 3. Also suppose that m and n are rational numbers. All of the coefficients and constants are real numbers. (b) The set of all negative integers. each of which has multiplicity 1. (c) the set containing 0. The result is (a) equal to 0. (b) the null set. (d) the set containing the null set. If we raise a to the mth power and then raise that quantity to the nth power. (c) There is one real solution with multiplicity 1. (e) undefined. When we factor that equation. (e) the universal set. and a second. because there are no real solutions at all. “Suppose that a is a nonzero number. we get the same result as if we ________. Fill in the blank to make the following statement true. (b) There are three different real solutions. 32.

FE-3. (b) Yes. It is called the associative law. 40. (a) |x| is six orders of magnitude larger than |p|. Based on the information in the drawing. (c) |x| is five orders of magnitude larger than |q|. 38. The quadratic equation stated in Question 36 has (a) two distinct real roots.554 Final Exam
36. and P. (d) Yes. (b) |x| is one order of magnitude smaller than |q|. and p are integers. (c) one imaginary root with multiplicity 2. Suppose somebody tells us that there’s a general law about how the terms in a subtraction problem can be grouped. (b) |x| is six orders of magnitude smaller than |p|. (e) two complex roots that are conjugates of each other. if m. 39. (a) |x| is one order of magnitude larger than |q|. According to that person. It is called the law of additive inverses. It is called the distributive law. what is it called? (a) This isn’t a legitimate law of mathematics. Suppose the numbers corresponding to these points are called x. then it is always true that (m − n) − p = m − (n − p) What can we say about this? Is the person right? Is this a legitimate law of mathematics? If so.
. (d) |x| and |q| have the same order of magnitude. (d) two imaginary roots that are additive inverses of each other. Q. (e) the order-of-magnitude relationship between |x| and |q| can’t be defined. (e) Yes. (c) Yes. n. It is called the commutative law. and p respectively. Figure FE-3 represents all the rational numbers in power-of-10 form. Consider the following quadratic equation: 2x 2 + 5x + 8 = 0 What is the discriminant in this equation? (a) 391/2 (b) −391/2 (c) 39 (d) −39 (e) ±j39 37. Based on the information in Fig. (b) one real root with multiplicity 2. Three points are shown on the lines: X. q.

(d) We can multiply both equations through by 2. we get the same result as if we ________.999 (d) b. and then add it to the top equation. and p and q are nonzero integers. If we multiply these two quantities. 44. (e) We can multiply both equations through by 0. Suppose a. Which one is the exception? (a) p + q = q + p (b) p + (−q) = (−q) + p (c) (−p) + q = q + (−p) (d) p − q = q − p (e) (−p) + (−q) = (−q) +(−p) 43. 42. and then add them.999 (c) a.caa / 9. Suppose that m and n are integers. What is the fractional equivalent of the endless repeating decimal 0. and then add it to the bottom equation. and then add them. Consider a m and a n.999 (e) None of the above 45.556 Final Exam
and −4x + y = −2 How can we add multiples of these two equations to make x vanish. then mn = pq. “Suppose that a is a nonzero number. with coefficients that are all integers? (a) We can’t. (e) None of the above.? (a) a. (c) We can multiply the bottom equation through by −1/2.abcaabcaabcaabca . then m /q = n /p. Which of the following statements is always true? (a) If m /p = n /q. (c) If m /p = n /q...” (a) divide a by (m + n) (b) multiply a by (m + n) (c) raise a to the power of (m + n)
.abc / 9. leaving us with a solvable first-degree equation in y alone. then mp = nq. (d) If m /p = n /q.aab / 9. Fill in the blank to make the following statement true. (b) If m /p = n /q. (b) We can multiply the top equation through by 2. all different. where m and n are integers. then mq = np.999 (b) c. All of the following equations except one are generally true for any two integers p and q. and c are single digits. b.bca / 9.

What numeral represents a quantity one less than 100? (a) 9F. (e) its numerator and denominator are both prime. (e) real numbers. −s /r} (c) X = {pq. (b) its denominator is “cleanly” divisible by its numerator.Final Exam
557
(d) raise a to the power of mn (e) raise a to the power of 1/(mn) 46. (c) rational numbers only. where p. (d) irrational numbers only. −rs} (e) We need more information to answer this 48. s /r} (b) X = {−q /p. r. (b) 7F. and s are all positive real numbers: (px + q)(rx + s) = 0 What is the solution set X for this equation? (a) X = {q /p. (e) There is no such numeral as 100 in the hexadecimal system. q. (d) its denominator is a product of primes. we get (a) a 2 + b 2 (b) a 2 − b 2 (c) 2a (d) 2b (e) 4ab 47. rs} (d) X = {−pq. (d) FF. (c) AF. We can tell right away that a fraction is in lowest terms if (a) its numerator is “cleanly” divisible by its denominator. so this question is meaningless! 50. Consider this general equation. (c) its numerator is a product of primes. 49. (b) integers only. Imagine that we’re working exclusively in the hexadecimal system. When we add a + jb to its conjugate.
. The commutative law for addition can be applied to finite sums of (a) natural numbers only.

We want to solve these functions as a two-by-two system. (b) One. (c) Two. the maximal domain and the co-domain are represented by the solid gray regions. (b) Drawing B. (e) We can’t answer this without more information. where all the coefficients and constants are real numbers. Figure FE-4 shows four mappings. 52. (d) Drawing D. (c) Drawing C. In each case. Imagine that we come across two cubic functions.
Figure FE-4 Illustration for Final Exam
Question 52.
A
B
C
D
. What is the minimum number of real solutions that such a system can have? (a) None. (d) Three.558 Final Exam
51. Which of these drawings illustrates the concept of a surjection onto the co-domain? (a) Drawing A. (e) None of them.

where x is the unknown and a is a constant? (a) x = 5 + a (b) x = 5 − a (c) x = a − 5 (d) x = −a − 5 (e) It can’t be determined without more information 57. (c) the rectangular system arranges the four quadrants in a more sensible way than the Cartesian system.Final Exam
559
53. (b) −2. (d) 2/7. (e) undefined. (d) −7/4. (c) −7/2. The x-intercept of the graph of the equation stated in Question 53 is (a) 1.
. a rectangular coordinate system often works better than a strict Cartesian coordinate system because (a) the rectangular system shows the true slopes of the lines or curves. (e) impossible to determine without more information. (c) 7/4. (e) undefined. but the Cartesian system does not. (c) −7/2. 55. Consider the equation 4x + 2y = −7. When graphing a two-by-two system to illustrate the real solutions. (b) −2. (b) the rectangular system shows negative as well as positive solutions. 56. The slope of the graph of this equation is (a) 2. The y-intercept of the graph of the equation stated in Question 53 is (a) 2. (b) −1. 54. (d) 2/7. What is the solution to the equation −x + a + 5 = 0. (e) the rectangular system can often provide a better pictorial fit than the Cartesian system for the range of values we want to show. where x is the independent variable and y is the dependent variable. but the Cartesian system can provide only approximate values. (d) the rectangular system can provide exact values for the solutions merely by observation. but the Cartesian system shows only positive solutions.

For any point that’s part of the graph of the original relation. In which quadrant is this point? (a) The first quadrant. FE-5.
. (c) The third quadrant. 60. one of the points is labeled P. What is the equation of this “point reflector” line? (a) y = 1 (b) x + y = 0 (c) x − y = 0 (d) x = 1 (e) None of the above 59. a scheme is shown for graphing the inverse of a relation. FE-5. (b) The second quadrant.560 Final Exam
58. In Fig. (b) Only if the variable can never become equal to 0. In Fig.” exactly the same distance away. Under what circumstances can we add the same variable to both sides of an equation and get another valid equation? (a) Never. (e) It is not in any quadrant.
y 6 4 Inverse Original 2
Original “Point reflector” line Inverse
x –6 Inverse –4 –2 –2 Original –4 –6 Original Inverse P 2 4 6
Figure FE-5 Illustration for Final Exam Questions 58
and 59. (d) The fourth quadrant. we can find its counterpart in the graph of the inverse relation by going to the opposite side of the “point reflector.

(d) a surjection. A mapping in which each element in the domain corresponds to one.Final Exam
561
(c) Only if the variable can never become negative. Which of the following statements is false? (a) If a real number k is a root of a cubic equation in the variable x. (c) If k is a real number and (x − k) is a factor of a cubic polynomial in the variable x. (b) a bijection. then (x − k) is a factor of the cubic polynomial. (b) If a real number k is a root of a cubic equation in the variable x. (d) All cubic equations have at least one real root. (e) Always. then k is a real root of the cubic equation. (d) Only if the variable can never become irrational. (e) onto. 64. 63. 61. (e) converting it into a cubic equation that turns out to be easier to solve than the original quadratic. Consider the following quadratic equation in binomial factor form: (x − j3)(x + j3) = 0 What is the solution set X for this equation? (a) X = {3} (b) X = {j3} (c) X = {−3} (d) X = {−j3} (e) None of the above
. Completing the square is a method of solving a quadratic equation for real roots by (a) squaring the left and right sides. then (x + k) is a factor of the cubic polynomial. (b) adding a constant to both sides in order to get the square root of a binomial on one side and a positive real number on the other side. (e) A cubic equation can have one real root and two other roots that are complex conjugates of each other. element in the range is called (a) a rejection. but only one. 62. (d) taking the square root of both sides. (c) an injection. (c) adding a constant to both sides in order to get the square of a binomial on one side and a positive real number on the other side. discovering the imaginary roots if any exist. as long as the coefficients and the stand-alone constant are all real numbers.

71828). When we say that a real number u is the natural logarithm of some other real number v. (b) u equals e to the vth power. A three-by-three linear system is consistent and not redundant if and only if (a) it has a single. We can see this because (a) the curve is not a straight line. (c) We can restrict the range to the set of reals larger than 1. (e) it has no solutions. We can see this because (a) there are values of x that map into more than one value of y. (b) it has two distinct solutions. unique solution. (d) it has infinitely many solutions. (c) Only if the variable can never become negative. we are in effect saying that (a) v equals e to the uth power. 70. (c) the relation is not one-to-one. (d) there are values of x that map into more than one value of y. (c) it has three distinct solutions. FE-6 is not a function of x if we think of it as a mapping from values of x to values of y. (d) We can restrict the range to the set of reals smaller than −1. (b) there no values of y that map into more than one value of x. 68. (e) v to the uth power equals e. (e) Always. (c) v equals u to the eth power. (b) the domain is not the entire set of real numbers. 67. (e) Any of the above. 66. and that e is Euler’s constant (an irrational number equal to about 2. How can we restrict the range of the relation graphed in Fig. (b) Only if the variable can never become equal to 0. Under what circumstances can we divide both sides of an equation by the same variable and get another valid equation? (a) Never. The relation graphed in Fig. FE-6 so that it becomes a function of x if we think of it as a mapping from values of x to values of y? (a) We can restrict the range to the set of non-negative reals. (d) u equals v to the eth power. (b) We can restrict the range to the set of negative reals. (d) Only if the variable can never become irrational. The relation graphed in Fig. FE-6 is a function of y if we think of it as a mapping from values of y to values of x.562 Final Exam
65. 69.
. (e) there no values of y that map into more than one value of x.

(c) the curve is symmetrical. −j10} (b) X = {10. 71. (c) somewhere between the smallest and the largest real roots of the equation. (d) the domain is the entire set of real numbers. (e) smaller than all the real roots of the equation.Final Exam y 6 4 (0. This tells us that our “test root” is (a) larger than all the real roots of the equation.0) 2 y =+ -2 x1/2 x –6 –4 –2 –2 –4 –6 2 4 6
563
Figure FE-6 Illustration for Final Exam Questions 68. (b) equal to the largest real root of the equation. (10 − j10)}
. −10} (c) X = {(10 + j10). The numbers in the last row alternate between positive and negative. We try a negative real-number “test root” and get a nonzero remainder. 72. and 70. State the solution set X for the quadratic equation x 2 + 100 = 0 (a) X = {j10. Suppose that the coefficients and constant in a polynomial equation are placed in a synthetic division array. (d) equal to the smallest real root of the equation. (e) the relation is a surjection.
69.

(c) Five. (e) w to the z power equals 10. (d) 816. What is the slope of this line? (Be careful! Note that the horizontal axis represents t. We want to solve these functions as a two-by-two system. (c) z equals w to the 10th power. (b) Four. (e) 1.564 Final Exam
(d) X = {j10} (e) X = ∅ 73. we are in effect saying that (a) z equals 10 to the wth power. 75. (b) w equals 10 to the zth power. Which of the following equations is true for all positive real numbers u and v ? (a) ln u + ln v = ln (u + v) (b) ln u + ln v = ln (uv) (c) ln u + ln v = ln (uv) (d) ln u ln v = ln (uv) (e) None of the above 74. Imagine that we come across two different functions.023. 77. FE-7. both of which are of degree larger than 2. (d) Infinitely many. (b) 589. and where all the coefficients and constants are real numbers. (e) We must have more information before we can answer this. and the vertical axis represents x !) (a) 1/2 (b) 1 (c) −1/2 (d) −1 (e) None of the above
. (d) w equals z to the 10th power. the straight line is the graph of a function where t is the independent variable and x is the dependent variable. 76. In Fig. When we say that the common logarithm of a real number w is equal to z. (c) 600. What is the maximum number of real solutions that such a system can have? (a) Three. The octal numeral 700 represents the same quantity as the base-10 numeral (a) 448.

(e) impossible to determine without more information. (c) No. (d) −7/3. (d) Yes. Is that a good idea? (a) No. 79. so we can reduce it to a quadratic that will be easier to solve. (e) impossible to determine without more information. (e) Yes. because one of the real roots is x = 0.Final Exam
x (1. but only because the leading coefficient is equal to 1. FE-7.
78. the x-intercept of the line is (a) 13/4. the t-intercept of the line is (a) −11/6. (b) 7/2.
. it will work fine. (c) −13/6. (c) 4. and 79. (d) 17/4. because we must always factor cubics into binomials to solve them.6)
565
t
(–4. 78. 80. (b) −2. FE-7. In the graph of Fig. because a cubic equation can never be divided through directly.–4)
Figure FE-7 Illustration for Final Exam
Questions 77. (b) No. Suppose we see the following cubic equation and we want to find its real roots: x 3 − 4x 2 + 7x = 0 We’re tempted to divide this through by x. In the graph of Fig.

Which of the following equations is not a quadratic in one variable? (a) x 2 = 2x + 3 (b) 1/x = x 2 − 7 (c) x 2 + 4x = 27 (d) x − 21 = −8x 2 − 22 (e) 6 + x = 2x 2 86. Within the set of real numbers.22.566 Final Exam
81. What is the original integer n? (a) 11 (b) −13 (c) 29 (d) −12 (e) 0 84. (d) intersect in a single straight line. We add n to 2/3 of itself. (d) is equal to the 99th root of −1. 85. (b) is equal to −1 to the 99th power. the base-(−1) logarithm of 99 (a) is not defined. or −1. For any three-by-three linear system to have a single. (e) is equal to (ln 99) to the −1st power. or approximately 0. the graphs of all the equations must (a) not intersect anywhere. Imagine an integer n. Which of the following is a first-degree equation in one variable? Here. unique solution. 82. or 1/99. What is the sum of 3/5 and 7/12 in lowest terms? (a) 71/60 (b) 10/17 (c) 60/71 (d) 17/35 (e) 73/69 83. and d are constants. (c) intersect in a single flat plane. c. (c) is equal to 99 to the −1st power. (e) intersect at a single point. or −1. We end up with −11. (a) 3x 2 + 5x − 5 = 0 (b) ax + bx 1/2 + cx 1/3 = d
. x is the variable. (b) intersect at the origin of Cartesian three-space. while a . b. and then we subtract 3/4 of n from that sum.

we can perform any of the following operations except one. (b) both axes have increments of the same size. (b) a composite number that cannot be divided by any other natural number except 1 without a remainder. In a rectangular coordinate plane. keeping the elements in the same order from left to right. A prime number is (a) a natural number that can be factored into a product of composite numbers. 88. and that can only be factored into a product of itself and 1. When we morph a matrix representing a three-by-three linear system. (e) a natural number that is the cube of some composite number. the change in value is directly proportional to the distance we move along that axis. considering x as the independent variable and y as the dependent variable? (a) y = (−2/3)x − 2 (b) y = (2/3)x + 3/2 (c) y = (3/2)x + 3 (d) y = (−3/2)x − 8/5 (e) More information is necessary to answer this
. (d) Add all the elements in any row to all the elements in another row. while keeping the elements of both rows in the same order from left to right. 90. (c) a natural number that is 2 or larger. This is another way of saying that (a) along either axis. What is the slope-intercept equation of line L. (e) the axes intersect at a right angle. and then replace the elements in either row by the sum. (c) Multiply all the elements in a row by a nonzero constant.Final Exam
567
(c) 5x − a + b 2 = 8 (d) x 3 + x 2 + x = 1 (e) x − x 1/2 − x 1/3 = a + b 87. taking care to keep the elements of both rows in the same order from left to right. 89. (e) Add a constant to all the elements in a row. both axes are linear. (d) linear equations always have graphs that fall exactly on one of the axes. (c) the origin is at the point (0. keeping the elements in the same order from left to right. (d) a natural number that is the square of some composite number. (b) Divide all the elements in a row by a nonzero constant. keeping the elements in the same order from left to right. Which one is the “illegal move”? (a) Interchange all the elements between two rows. Figure FE-8 is the graph of a two-by-two linear system. 0).

or d to equal 0 at any time. b.Final Exam
569
(c) x = (−5/4)y − 2 (d) x = (−5/4)y + 2 (e) More information is necessary to answer this 94. (e) The fifth statement is wrong. (b) We cannot allow b. where a. (c) a parabola. We cannot. c. c. in general. Which of these statements fully states those conditions? (a) We cannot let b. • We can multiply both sides by the same quantity. (c) We cannot allow a to equal 0 at any time. (d) The fourth statement is wrong. (b) The second statement is wrong.” • We can add the same quantity to both sides. and d all equal 0 at the same time. • We can subtract the same quantity from both sides. (d) a circle. and how can it be corrected? • We can reverse the left and right sides only if we change the inequality to “larger than or equal. but if we multiply both sides by a negative quantity. Consider the following first-degree equation in the variable x. Something is wrong with one of these claims. (e) None of the above. Here are several things we can do to “smaller than or equal” statements and still have valid statements. the equation 2x + 4y − 6z = 7 represents (a) a straight line. Which claim is wrong. add one “smaller than or equal” statement to another.
.” 95. and f are constants: −3a + x /(bcd ) = 24ef This equation has meaning only under certain conditions. We can only add the same quantity to both sides of a “smaller than or equal” statement if that quantity is positive. e. (c) The third statement is wrong. It works if the quantity is nonnegative. c. (b) a flat plane. 96. • We can add one statement to another. We can never reverse the left and right sides of any inequality. (a) The first statement is wrong. In Cartesian three-space. we must change the relation to “larger than or equal. d. We can only subtract the same quantity from both sides of a “smaller than or equal” statement if that quantity is negative.

Assuming the condition in Question 96 has been satisfied. (c) we always get a rational number. Imagine two straight lines in this Cartesian plane. (d) we always get a negative number. (c) Neither function has an inverse that is also a function.
.–6)
Figure FE-9 Illustration for Final Exam
Questions 99 and 100. and the other passing through lines R and S. FE-9. 99. When a quantity is raised to the −1 power.
y
P = (0. (b) we always get the additive inverse of the quantity. (d) The system is redundant. 97. (b) The system is inconsistent. (e) We cannot allow f to equal 0 at any time. Now think about the two-by-two system of linear equations represented by these lines. What can we say about this system if we consider both of the equations as functions of x ? (a) The system has one solution. (e) we always get 1.4) R = (5.2) x Q = (–4. one passing through points P and Q. but the other function does not.–4) S = (1. what is the solution of this equation in terms of the constants? (a) x = 3abcd + 24bcdef (b) x = abcd /3 + bcdef /24 (c) x = 3abcd − 24bcdef (d) x = abcd /3 − bcdef /24 (e) More information is needed to answer this 98. (e) One of the functions has an inverse that is also a function. except when the quantity equals 0. (a) we must be sure the quantity can never equal 0. Refer to Fig.570 Final Exam
(d) We cannot allow e to equal 0 at any time.

(b) two solutions. The matrix shown in Question 101 contains enough information so that we can infer the solution to the linear system it represents. Suppose we are trying to solve a three-by-three linear system using matrices. z)? (a) (3. b. −4. 101. FE-9. −36. 5) (e) (12. (d) unit diagonal form. imagine two straight lines. and d are constants. (c) three solutions. c. 0) (b) (6. We are sure we haven’t made any mistakes along the way.Final Exam
571
100. −3. 12. 102. (c) diagonal form. and z are variables. and x. and the other passing through lines Q and R. Now suppose that the following matrix represents this system:
2 0 0 0 −3 0 0 0 5 6 12 0
This matrix is in (a) linear form. Consider the two-by-two system of linear equations represented by these lines. (d) four solutions. This time. (e) redundant form. We come up with this matrix:
7 1 −15 7 1 −15 7 1 −15 28 4 −60
. each of which is in the following form: ax + by + cz = d where a. y. 5) (d) (8. y. How can that solution be expressed as an ordered triple of the form (x. 0) (c) (2. Imagine a three-by-three linear system of equations. Look again at Fig. This system has (a) one solution. 9. (b) dependent form. (e) infinitely many solutions. 0) 103. one passing through points P and S.

(e) infinitely many solutions. (b) The first expression represents a complex (but not real) number. The second expression represents a number that is neither pure real nor pure imaginary. The second expression represents a real number.572 Final Exam
We can conclude that the original three-by-three system has (a) no solutions. (c) two solutions.
jb j6 j4 j2 a -6 -4 -2 -j 2 Circle centered at origin -j 4 -j 6 2 4 6 (0. (e) The first expression represents a complex number that is neither pure real nor pure imaginary. The second expression represents a complex (but not real) number. FE-10. Are the expressions 2 + j 2 and 2 + j 2 different? If so. The second expression represents a pure imaginary number. 105.j3) representing 0 + j3
Figure FE-10 Illustration for Final Exam Question 105. (d) The first expression represents a pure imaginary number. (b) They are all pure imaginary numbers. what do all the numbers corresponding to points on the circle have in common? (a) They are all pure real numbers. In Fig. (d) three solutions.
. (b) one solution. but they actually represent the same number. 104. how? (a) These two expressions look different. (c) The first expression represents a real number.

we know that (a) the solution values for x and y are both positive. Which of the following statements is false? (a) j 7 = j 3 (b) j 8 = j 4 (c) j 12 = j 8 (d) j 16 = j 12 (e) j 21 = j 18 108. Here is a little mathematical verse: For every x. 106. (b) It’s a number that we can add to a given number to produce a sum equal to the original number. (e) the solution value for x is the additive inverse of the solution value for y. What is meant by the term additive identity element? (a) It’s a number that we can add to a given number to produce a sum of 0. y. How can we write this in mathematical symbols? (a) (∀ x. From this information. Imagine a two-by-two linear system in variables x and y. y. y. (d) the solution values for x and y are both negative. and the solution value for y is positive. then x is smaller than z. y. and the solution value for y is negative.
. (e) They are all equal to the square root of −9. (c) the solution value for x is negative. (b) the solution value for x is positive. z) : [(x ≥ y) & (y ≤ z)] ⇒ (x < z) (d) (∀ x. z) : [(x ≤ y) & (y ≤ z)] ⇒ (x < z) (b) (∀ x.Final Exam
573
(c) They all have absolute values equal to 3. and z : If x is smaller than y and y is equal to z. Suppose the graphs of the equations are straight lines that intersect at a point in the second quadrant of a Cartesian coordinate system where y is the dependent variable. z) : [(x = y) & (y < z)] ⇒ (x < z) 109. 107. (d) They are all complex conjugates. z) : [(x < y) & (y > z)] ⇒ (x < z) (e) (∀ x. y. y. z) : [(x < y) & (y = z)] ⇒ (x < z) (c) (∀ x.

how many real zeros does the quadratic function have? (a) More than two. (b) It is a complex (but not real) number. (d) It is a negative real number.
. (c) One. Based on the information shown. (d) None. (e) range. (c) It is a positive real number. Figure FE-11 is graph of a quadratic function where x is the independent variable. (b) Two. 6/7). The essential domain of a mapping is always a subset of the (a) maximal domain. (b) (r. Based on the information shown in Fig. (c) (r. s) = (1/4. 7). (e) We need more information to answer this. (b) codomain. 110. (d) dependent variable. (e) It’s a number that we can add to a given number to produce a sum equal to twice the original number. (d) There are infinitely many solutions. (d) It’s a number that we can add to a given number to produce a sum of 1. s) = (−8/7. (e) We need more information to answer this. 113. s) = (−2. (c) independent variable. 111. (e) There is no solution.574 Final Exam
(c) It’s a number that we can add to a given number to produce a sum equal to the negative of the original number. Consider the following pair of equations as a two-by-two linear system: 2r = 6s − 8 and −7r = −21s + 28 What is the solution to this system? (a) (r. what can we say about the coefficient of x 2 in the polynomial standard form of the function? (a) It is an imaginary number. 112. −3/4). FE-11.

13. 5}. 19. we can determine that one of the following numbers is prime. (b) The set {1. What is the intersection of the sets A = {1. 3}. (b) The set {1. 5}. 2. 3} and B = {3. 17. 3. 3}. 4. 3. Which one? (a) 407 (b) 423 (c) 437 (d) 457 (e) 473 117.
114. (d) The set {1. (c) The set {3. and 23. (c) The set {3. 5}. 4. 4. 5}? (a) The set {3}. 2. 115. 4. 2. nonrepeating decimal.
. 5. 5}. 4. (d) The set {1. 3} and B = {3.0)
Figure FE-11 Illustration for Final Exam
Questions 112 and 113. 2. (b) a ratio of 1 to an integer. 11. 116. Using this information along with a calculator. (e) The null set. 5}? (a) The set {3}.Final Exam
y
575
Graph of a quadratic function
x
Origin point (0. 7. (e) The null set. What is the union of the sets A = {1. The first few primes are 2. 2. 2. 3. 4. A rational number can always be expressed as (a) an endless.

000 times as large by (a) adding 10. 122. (a) One..305 123. 124. (b) Two. Suppose someone gives us a list of the first few elements of an infinite set S. We can make the absolute value of this integer 10.096 (b) 8. and assures us that the numbers keep doubling as we move down the list: S = {2. 8. .} Which of the following is not an element of S ? (a) 4. 64.048. (c) Three.. (b) Two. Based on the information shown. how many real zeros does the quadratic function have? (a) More than two.
. 125. (e) We need more information to answer this. (d) It is a negative real number. (b) adding four ciphers to the left-hand end of the digit sequence. FE-12. (d) increasing the left-most digit by 4. (e) Infinitely many.Final Exam
577
What is the largest number of elements that the solution set of such a system can have? Assume that the two equations are not identical.576 (e) 4. Figure FE-12 is graph of a quadratic function where x is the independent variable. (e) inserting a decimal point four digits in from the right-hand end of the digit sequence. and are not some constant multiple of each other. 32.194.192 (c) 16. (c) It is a positive real number. 4.384 (d) 1. Based on the information shown in Fig. 16. (d) Four. (b) It is a complex (but not real) number. what can we say about the coefficient of x 2 in the polynomial standard form of the function? (a) It is an imaginary number. (c) adding four ciphers to the right-hand end of the digit sequence.000. Consider an integer expressed as a digit sequence without any other symbols (except a minus sign if the integer is negative).

. our goal is to find a “test root” that produces (a) a remainder of 0 at the end of the process. 23}. y. 1. Imagine four nonzero real numbers w. (d) alternating positive and negative reals in the last line. (d) {0.. (c) all positive reals in the last line.0)
Figure FE-12 Illustration for Final Exam
Questions 124 and 125. (d) z /x = w /y. (e) None of the above.. 2. . (e) We need more information to answer this... (e) all negative reals in the last line. 126.
(c) One. (b) all 0s in the last line. 1. (e) {24}. 3. 23. . . 3.578 Final Exam
y
Graph of a quadratic function
x
Origin point (0. (c) {0.. (c) w /x = y /z. the natural number 24 is (a) {1. 22. 128.
. 22.. we can conclude that (a) w /y = x /z. 21. (d) None. From this. 21. x. (b) {1. When using synthetic division in an attempt to factor a binomial out of a polynomial. 2. 24}. 22. (b) w /z = x /y. and z such that wx = yz. 22... 127. 23}. 23. According to the set-based number-building scheme. 2. 24}.. 2. ...

None of the numbers in the last row are negative. and the other solution is an ordered pair of negative reals. Suppose we consider these two functions as a system. (c) somewhere between the smallest and the largest real roots of the equation. Suppose that the coefficients and constant in a polynomial equation are placed in a synthetic division array. (a) both solutions are ordered pairs of positive real numbers. (e) smaller than all the real roots of the equation. and we find two solutions. (b) both solutions are ordered pairs of negative real numbers. (d) equal to the smallest real root of the equation. The origin (0. (c) one solution is an ordered pair of positive reals. Which of the following fractions is in lowest terms? (a) 7/91 (b) 57/3 (c) −23/115 (d) −29/17 (e) None of the above 131. (b) equal to the largest real root of the equation. We try a positive real-number “test root” and get a nonzero remainder. The curves do not intersect anywhere.Final Exam
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129. Figure FE-13 shows the graphs of two quadratic functions in a real-number rectangular coordinate plane. This tells us that our “test root” is (a) larger than all the real roots of the equation. Based on the information shown. 130.
y
x
Figure FE-13 Illustration for Final
Exam Question 131. 0) is where the x and y axes intersect.
.

If a number decreases by a factor of 100. but can’t be both. and the other solution is an ordered pair of pure imaginary numbers. (c) We can factor the equation into binomials with integer coefficients and integer constants. (e) We can use a computer program to graph the function produced by the polynomial. (c) increases by a factor of exactly 2.580 Final Exam
(d) one solution is an ordered pair of reals. (e) increases by a factor of the square root of 100. (d) An odd number times 2 is always a natural number. or approximately 9.000. and then derive the roots from those binomials. or exactly 10. The origin (0. What can we do to find. Figure FE-14 shows the graphs of two quadratic functions in a real-number rectangular coordinate plane. and then find those roots using the rational roots theorem. (e) increases by a factor of the square root of 10. 135. 134. make that trinomial into a quadratic. 133.2. Which of the following statements concerning natural numbers is false? (a) A number must be either even or odd. 132.2. (b) We can factor the equation into the nth power of a trinomial (where n is the degree of the equation). or at least approximate. If a number increases by a factor of 10.6. (c) An even number divided by 2 is always a natural number. then its natural logarithm (a) decreases by a factor of ln 100. (b) decreases by a factor of exactly 2. (b) decreases by a factor of exactly 4. or approximately 4. (b) An even number plus 1 is always odd. and then solve that equation. make that binomial into a first-degree equation. The
.000.000.6. (e) both solutions are ordered pairs of complex (but not real) numbers. those roots? (a) We can factor the equation into the nth power of a binomial (where n is the degree of the equation). (d) We can use synthetic division to find the upper and lower bounds of the real roots.000. and then solve that equation with the quadratic formula. (d) increases by a factor of ln 100. or approximately 4. 0) is where the x and y axes intersect. 136. (d) increases by a factor of ln 10. and we’re told that all the roots are irrational numbers. (e) An odd number divided by 2 is always a natural number. (c) increases by a factor of exactly 4. or exactly 100. Suppose we want to solve a higher-degree polynomial equation. and then use the computer to approximate the zeros of that function. or approximately 9. then its common logarithm (a) decreases by a factor of ln 10.

and the other is an ordered pair of complex (but not real) numbers. In any such ordered pair. 138. Consider the following equation. (c) there is a single solution with multiplicity 2. the second number represents a value of the (a) codomain. Suppose we consider these two functions as a system. we get (a) a 2 + b 2 (b) a 2 − b 2
. 137. it is an ordered pair of pure imaginary numbers.5) to show specific examples of this function. and the other is an ordered pair of pure imaginary numbers. which represents a function of the variable x : y = x 2 + 3x + 1 We can write down ordered pairs such as (0.Final Exam
y
581
x
Figure FE-14 Illustration for Final
Exam Question 136. (e) there are two solutions. (d) there are two solutions. one is an ordered pair of reals. When we multiply a + jb by its conjugate. (e) essential domain. it is an ordered pair of complex (but not real) numbers. one is an ordered pair of reals. (c) bijection.1) or (1. (b) there is a single solution with multiplicity 2. (b) dependent variable.
graphs intersect at a single point. (a) there is a single solution with multiplicity 2. Based on the information shown. it is an ordered pair of real numbers. (d) inverse.

(b) We can decrease the value of c1. and a1 and a2 are both nonzero. (c) there is a single solution with multiplicity 2. one is an ordered pair of reals. Figure FE-15 shows the graphs of a linear function and a quadratic function in realnumber rectangular coordinates. (d) there are two solutions. The origin (0. The graphs intersect at a single point. and the other is an ordered pair of complex (but not real) numbers. and the other is an ordered pair of pure imaginary numbers. (e) there are two solutions. (b) there is a single solution with multiplicity 2.
. 0) is where the x and y axes intersect. it is an ordered pair of reals. it is an ordered pair of complex numbers. 0) is where the x and y axes intersect. Figure FE-16 shows the graphs of two quadratic functions in real-number rectangular coordinates. leaving everything else the same. The origin (0. 145. (a) there are two solutions. both are ordered pairs of real numbers. How can we change this system into one that we can be certain has no real solutions? (a) We need more information to answer this.Final Exam
y
583
x
Figure FE-15 Illustration for Final Exam
Question 144. The functions are f (x) = a1x 2 + b1x + c1 and g (x) = a2x 2 + b2x + c2 where all the coefficients and constants are real numbers. The graphs intersect at two points. one is an ordered pair of reals. Based on the information shown.
144. Suppose we consider these two functions as a system.

in general. 146. (e) It will cause us to unwittingly subtract an equation from itself. She tells us to: • • • • • Take the first equation and solve it for x in terms of y and z. Make a two-by-two system in y and z from the first two steps.
. There’s something else wrong with the process outlined in Question 146. Solve that system. leaving everything else the same. Suppose someone claims that she has come up with a way to solve a three-by-three linear system in variables x. (e) We can reverse the sign of a2.
This process will not. (d) The process does not employ matrix morphing.
(c) We can increase the value of c1. What’s that? (a) The third “bulleted” step will not actually give us a two-by-two system. leaving everything else the same. Substitute the solutions for y and z back into the first equation and solve it for x. work to solve a three-by-three system. leaving everything else the same. y. (e) The process can produce a solution only if the system is inconsistent.584 Final Exam
y
g (x)
x
f (x)
Figure FE-16 Illustration for Final Exam
Question 145. (d) We can decrease the value of c2. and z. (b) The two-by-two system derived in the third step will be inconsistent. Take the second equation and do the same thing. (c) The process does not use the addition method at any point. (d) It will cause us to unwittingly divide by 0. 147. (b) The process completely ignores the third equation. (c) There is no way to derive an expression for x in terms of y and z. Which of the following constitutes a fatal flaw in our friend’s scheme? (a) The process can work only if the system is consistent.

it means that (a) the equation has two different real roots. (c) None. (b) a function. If the discriminant in a quadratic equation is a negative real number.Final Exam
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148. and two. 150. (d) the equation has two different imaginary or complex roots. (a) None. What are the smallest and largest numbers of elements that the solution set of such a system can have? Assume that the two equations are not identical. b1. and are not some constant multiple of each other. (e) transitive. (e) the equation has no roots at all. and infinitely many. 149. (b) the equation has one real root with multiplicity 2. Consider the following pair of equations as a two-by-two system: y = a1x + b1 and y = a2x + b2 where a1. a2. and infinitely many. (e) Two. (d) a subset of its co-domain. (c) one-to-one and onto. (b) None. and infinitely many. and b2 are real numbers. and neither a1 nor a2 are equal to 0. (d) One. (c) the equation has one imaginary root with multiplicity 2. real or complex. and one.
. A mapping is a bijection if and only if it is (a) reflexive.

006. 6 would be written out in words as three hundred two quadrillion.000.070.000 + 1.611. six million. You can add as many ciphers as you want to the left of the digit 3 in the answer to the previous problem. which is equivalent to a thousand million. seventy trillion. add three ciphers to its right and then reposition the commas.. one hundred ten looks like this as a decimal numeral: 302. and it does not change the value of the number it represents.149.. one hundred ten thousand.900..110 You can keep “attaching ciphers” forever in the left-hand direction! 6.000 + 6 = 2.701.207.000 7.588 Worked-Out Solutions to Exercises: Chapters 1 to 9
(c) MMIX = 1. as follows.. Based on the numeral shown for the answer to Prob. seventy billion. 4 above. A mathematician might write it as . and a trillion is a million million.006. 5.302. six thousand.. tally the values of each digit in the decimal system and then add them up.110 Note the placement of the commas and the ciphers. one hundred forty-nine million. Also note that we use the American billion. 4.000 + 9 = 2. Starting with the numeral shown for the answer to Prob.100 If you want to make a number a hundred times as large.070. That’s based on the United States terminology where a billion is a thousand million. You can make any number in the decimal system ten times as large by adding one cipher to its right and then repositioning the commas. The number in the final answer to Prob.000. that gives you 302. 8. To solve this problem. • One times one gives you 1 • One times two gives you 2 • Zero times four gives you 0
. one hundred forty-nine billion.000 + 1. add two ciphers to its right and then reposition the commas. The number three hundred two trillion.490..110. you would get 3.009 (d) MMVI = 1..149.014. that gives you 30. This is a slightly whimsical problem.070.149.061.000 If you want to make a number a thousand times as large. Starting with the numeral shown for the answer to Prob.006.020.006 4. 4.

and you always end up with the same set A. Look at an example. That’s tricky and hasn’t been covered in this chapter. but the null set is not itself nothing. 9. The null set is a subset of itself. {∅}. If you let the written word “nothing” actually stand for nothing. That gives you the null set. 8. that’s tricky and hasn’t been discussed. First. you can add nothing.Chapter 2
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• One times eight gives you 8 • Zero times sixteen gives you 0 • One times thirty-two gives you 32 The decimal-numeral equivalent is therefore 1 + 2 + 0 + 8 + 0 + 32 = 43. An empty bank account is a perfectly valid account unless somebody closes it. You can “cheat” and use Table 1-2. plus three more. The null set contains nothing. but
. The null set lacks elements. eleven is represented by B. That means you can define quantity forty-three as the octal numeral 53. 2. You can say that it contains nothing as an element! If you have a certain set A with known elements. there are at least three ways to solve this problem. 2. like an empty bank account lacks money. You can also add up the values in octal arithmetic. To do it manually. the decimal equivalent as derived in the solution to Prob. note that forty-three. Then consider the set containing the null set. To do it manually. 3} Keep in mind that a subset is not the same thing as a set element. so we won’t show it here. note that forty-three is equal to two times sixteen. 10. That means you can define the quantity forty-three in hexadecimal form as the numeral 2B. 3} so therefore ∅ ⊆ {1. that is. plus eleven more. 1. then ∅ = { } = {nothing} and {nothing} ⊆ {nothing. As with the decimal-to-octal conversion.
Chapter 2
1. take nothing and make it an element of a set. The null set is a subset of any set. although not a proper subset of itself. This is a legitimate mathematical thing. Again. so we won’t do it here. 2. There are at least three ways to solve this problem. This is the quantity you commonly imagine as forty-three. is equal to five times eight. The last method is to add up the values in hexadecimal arithmetic. You can build up an infinite number of sets if you start out with nothing. In hexadecimal notation. You can “cheat” and use Table 1-2.

B and C. The universal set (call it U ) is a subset of itself. forever You started with nothing. you should be able to sense where this is leading: ∅ {∅} {{∅}} {{{∅}}} {{{{∅}}}} {{{{{∅}}}}} ↓ and so on. represented by the small dark-shaded triangle.” It can work well in a courtroom. is common to (that is. But U is not a proper subset of itself. is fully shared by sets B and D. is a proper subset of the set of all whole numbers. Whenever two regions are entirely separate. namely {{∅}}. 3. Both of these sets have
. You can write this as P=A∩C Set Q. The set of all even whole numbers. 6. and you have turned it into an infinite number of mathematical objects! A variation of this idea is handy for “building” numbers. too. irregular four-sided figure. U is the set of all entities. and the intersection of those sets is the null set. Therefore. In Fig. the only null-set intersection pairs are A∩D=∅ B∩C=∅ C∩D=∅ 5. which you can write as Q=B∩D 4.590 Worked-Out Solutions to Exercises: Chapters 1 to 9
it’s not the same thing as the null set. You can see from Fig. That’s impossible. 3. region Q represents the intersection of sets B and D. set P. If U were a proper subset of itself. real or imaginary. dark-shaded. Remember. Therefore. W. That’s trivial. 2-6 that the only pairs of regions that don’t overlap are A and D. shared by) sets A and C. it contradicts the very definition of U ! This little argument is an example of a tactic called reductio ad absurdum (Latin for “reduction to absurdity”) that mathematicians have used for thousands of years to prove or disprove “slippery statements. because any set is a subset of itself. It can be the sole element of another set. then the sets they represent are disjoint. region P represents the intersection of sets A and C. 2-6. as you’ll see in Chap. represented by the small. Therefore. and C and D. By now. There are plenty of examples that will work here. Weven. then there would be some entity that did not belong to U.

one at a time. that belong to both sets.. You should get ∅ {1} {2} {3} {1.} This is one of the strange things “infinity” can do. but they can count only once).” and the resulting set is exactly the same “size” as the original set. 2} {1. 3}. . because when you perform the divisions. you get {0. 1/6. 10/2. Amazingly enough. but only those elements.. which is a subset of any set.} It’s not hard to see that set G contains all those elements.} G = {1. You can take away every other element of a set that has infinitely many elements and that can be written out as an “implied list. 7. 3}
. 3. 8/2. you can pair the elements of both sets off one-to-one! The mechanics of this goes a little beyond the scope of this chapter.Chapter 2
591
infinitely many elements. 1/8. 2. 1/5. 4. and then write down the first few elements of the resulting set. 3} {1. you get the same set A (with certain elements listed twice. 2. 1/2. 3} {2.} But that’s exactly the same as W. .. . They are 1. you get {0/2. 1. or both. 4/2.. Then start the list of subsets by putting down the null set. 1/2. 2. That means set A contains precisely those elements that belong to one set or the other. 2. 1/3.. so A∪G=A 8. Then assemble all the sets you possibly can.. 2/2. 2. Let’s write out the sets again here as “implied lists”: A = {1.. isolate all the individual elements of the set {1. 1/4. 5. 1/32. First. . but you can get an idea of how it works if you divide every element of Weven by 2. Therefore A∩G=G If you start with set A and then toss in all the elements of G. When you do that. 6/2. and 3. 3.. using one or more of the elements 1. 1/4. and list them. 1/16.

{2. The number 6 is divisible by 3 without a remainder: 6/3 = 2 Of course. multiplication by 3 always produces an odd number: 3×1=3 3×3=9 3 × 5 = 15 3 × 7 = 21 3 × 9 = 27
. {3}}}. 2. {2. 3}} is ∅ {1} {{2. {3}}} has two elements: the number 1 and the set {2. 3}} 10. {2. you always get an odd number as the product. Therefore. {3}}. {2. 3} down and have it remain an element of the original set {1. you’ve made a mistake somewhere. (Proving that was one of the “challenge” problems in this chapter. If they’re different. {3}}} {1. In the number-building systems described in this chapter. the set of all possible subsets of {1. {2. {3}}} When writing down complicated sets like these. No. The set {1.) For the first few odd numbers. 3}} {1. nothing doesn’t represent any number. the set of all possible subsets of {1. is even. 3. {3}}} is ∅ {1} {{2. 3}}. {2. 3}} has only two elements: the number 1 and the set {2. This is the ultimate trivial question.
Chapter 3
1. 3}. There are plenty of other even numbers that can be divided by 3 leaving no remainder. The set {1. Therefore.592 Worked-Out Solutions to Exercises: Chapters 1 to 9
9. You can’t break {2. {2. 2. {2. {3}} down and have it remain an element of the original set {1. When you multiply an odd number by 3. the quotient here. The reason for this is similar to the reason why any even number times 7 gives you another even number. You cannot break {2. always be sure that the number of opening braces is the same as the number of closing braces.

You should get 32 with a decimal point and some digits. 29. 31} Now. 13. 7. it looks like one of the following: ______1 ______3 ______5 ______7 ______9 where the long underscore represents any string of digits you want to put there. As things turn out.081. Therefore. then 901 is prime. list the set all the primes less than or equal to 33. 5. 19. However large it is. 901 is not prime. 4. 23. Using Table 3-1.081 therefore has prime factors of 23 and 47. The number 1. The odd number on top. 7.
. then 901 is not prime. You will see that 1. 31} Divide 1. 29. start by taking its square root using a calculator. First take its square root using a calculator. 5.Chapter 3
593
You can prove that multiplying any odd number by 3 always gives you an odd number if you realize that the last digit of an odd number is always odd. Think of an odd number.081 = 23 × 47 Both of these factors are prime. If the number on the bottom is 3. you can list the set all the primes up to and including 31.081 is divisible by 23 without a remainder. divide 901 by each of these numbers. That set is {2. 3. If you fail. To find the prime factors of 1. 11. 23. 3. Now think of “long multiplication” by 3. then the last digit in the product must be 3. 17. or 7 (the second digit in 27) respectively. 3. 11. or 9. Round this up to the next whole number. getting the last digit of the product. 901 is divisible by 17 without a remainder. 1 (the second digit in 21). Using Table 3-1. which you are multiplying by the number on the bottom. Therefore. Round this up to 33. 5. 5. 5 (the second digit in 15). as you can see by looking at Table 3-1. 9. which is 31. any odd number times 3 is always odd. You always start out by multiplying the last digits of the two numbers together. Remember how you arrange the numerals on the paper and then do the calculations. You’ll get 30 with a decimal point and some digits. 17. That set is {2. try to break it down into a product of prime factors. You get 1. 19. 13. 7. must end in 1. If you succeed in doing that.081 by each of these numbers. To figure out whether or not 901 is prime.

you’ll see that 2. Remember from basic arithmetic that whenever you multiply a positive number times another positive number. 19. all equal to 13. Then 0 would also be composite. In the traditional sense. . When a natural number is multiplied by itself and then the result is multiplied by the original number again.197 into a product of primes (again using the process you did to solve Probs. 3. You can pair the set of all natural numbers one-to-one with the set of all integers like this. −3. 11. because they can’t be broken down into factors from the traditional set of primes {2. then 1 times any other prime would be equal to that same prime. the result is always positive.}.197 is a perfect cube of a prime.197 = 13 × 13 × 13 It has three prime factors. That means no negative number can be composite if we stick to the traditional definition of a prime number. and you always get 0. forever
. you start with 0 and proceed through the positive and negative integers alternately. one after the other. . with natural numbers on the left and integers on the right: 0↔0 1↔1 2 ↔ −1 3↔2 4 ↔ −2 5↔3 6 ↔ −3 ↓ and so on.. this “implied one-ended list” will eventually get to it.. because you can multiply 0 by any prime you want. 5 and 6). 5. 5. When you factor 2. 13. When you break the number 841 into a product of primes using the same process as you did to solve Prob. An even more serious problem occurs if we let 1 be called prime. 8. no matter how large positively or negatively. 3-5. 9. 1.} If you pick any integer. In Fig. you’ll see that 841 = 29 × 29 It is a perfect square of a prime. 3. this way: Z = {0. 10. they can’t be composite either. You can create an “implied one-ended list” of the entire set Z of integers starting with 0. 7.. 7. making every prime number composite! That’s why mathematicians generally refuse to call 0 or 1 prime numbers. −2. 17. But when we don’t allow them to be prime. −1. the product is a perfect cube.. If that were true. all the primes are positive because they are all natural numbers larger than 1.594 Worked-Out Solutions to Exercises: Chapters 1 to 9
6. Any composite number can be factored into a product of primes. The number 2. Suppose 0 were defined as prime. 2.

either of these equations will hold true for all possible integers.
If positive change represents warming. You can then substitute a for b and get a−a=a−a You could also substitute b for a. This works whenever b = a. and cooling as a negative temperature change (although there is no technical reason why you couldn’t do it the other way around). a and b can be any integers you want. Then you get (a − b) + 0 = a − (b + 0) which simplifies to a−b=a−b That equation holds true for any a and b you could possibly choose. −6. In order to see why. −1.596 Worked-Out Solutions to Exercises: Chapters 1 to 9
4. January 1999 averaged 3 degrees warmer than January 1998. January 2001 averaged the same temperature as January 2000. 0. getting b−b=b−b Obviously. This always works if c = 0. 5. there was no difference in the average temperature for January 2005 as compared with January 1998. January 2002 averaged 7 degrees warmer than January 2001. 7. The values of a and b don’t have to be the same. January 2004 averaged 2 degrees warmer than January 2003. January 2000 averaged 6 degrees cooler than January 1999.
. 2. You add these integers to discover the change in average temperature between January 1998 and January 2005: (−5) + 2 + (−1) + 7 + 0 + (−6) + 3 = 0 In Hoodopolis. With that restriction. Here are the year-by-year records once again for reference: • • • • • • • January 2005 averaged 5 degrees cooler than January 2004. you can start with the original equation: (a − b) + c = a − (b + c) “Plug in” 0 for c. then going backward in time you have year-toyear changes of −5. 6. and 3 degrees. January 2003 averaged 1 degree cooler than January 2002. It makes sense to consider warming as a positive temperature change.

Then you get (a − b) − 0 = a − (b − 0) which simplifies to a−b=a−b Here are a couple more extra-credit exercises.
Statements a+b+c+d a + (b + c + d ) a + (d + c + b ) (d + c + b) + a d+c+b+a Q. but you let a and c be any integers? You’re on your own! 7. and d are added. Solution to Prob. and get the same sum. c. but you let b and c be any integers? Does it work when b = 0. each statement is equal to every statement that came above it. This proves that for any four integers a. b.E. See Table A-1.Chapter 4
597
Here are a couple of extra-credit exercises if you want a further challenge. Follow each statement and reason closely so you’re sure how it follows from the statements before. 4. and d. This always works if c = 0. This S/R proof shows how you can reverse the order in which four integers a. Does the original rule work when a = 0. Does the original rule work when a = 0. Here’s the original equation: (a − b) − c = a − (b − c) “Plug in” 0 for c. but you let b and c be any integers? Does it work when b = 0. b. Let’s start with the original expression. Then a and b can be any integers you want. As you read down the left-hand column. We begin with this: (a + b + c) + d
Table A-1. c. Reasons Begin here Group the last three integers Result of the “challenge” where it was proved that you can reverse the order of a sum of three integers Commutative law for the sum of a and (d + c + b) Ungroup the first three integers Mission accomplished
.D. 8 in Chap. and then morph it into the second one. but you let a and c be any integers? Have fun! 8. a+b+c+d=d+c+b+a 9.

Table A-2. generalized b + (−b) = 0 and c + (−c) = 0 so the b’s and c’s “cancel out”
. In this particular situation. 10. (a + b + c) + d = a + (b + c + d ) If this seems like a trivial exercise. keep in mind that you’re getting your brain into shape to do more complicated tricks some day. replace original parentheses with brackets for temporary clarification Get rid of outer sets of brackets. As you read down the left-hand column. Solution to the first part of Prob. each statement is equal to every statement that came above it. 4. and c. c. The second expression can be simplified as shown in Table A-3. These are called S/R derivations.
Statements (a + b − c) + (a − b + c) [a + b + (−c)] + [a + (−b) + c] a + b + (−c) + a + (−b) + c a + a + b + (−b) + c + (−c) a+a Reasons Begin here Change subtractions into negative additions. b. b. which breaks the process down into statements and reasons. they are not necessary in straight sums Commutative law for addition. after all). the two original expressions turn out to be equivalent for all integers a. and call the “package” e. to get a + (b + c + d ) That’s the result we want. proving that for any four integers a. The first expression can be simplified as shown in Table A-2.598 Worked-Out Solutions to Exercises: Chapters 1 to 9
We can “zip up” the sum b + c. They’re not intended as proofs. but to show how an expression can be morphed into another expression. and d. 10 in Chap. Now our expression looks like this: (a + e) + d The associative law for addition tells us that we can rewrite this as a + (e + d ) Now let’s “unzip” e so it turns back into the sum b + c (which it always has been.

4. If you start with any integer and keep multiplying by −3 over and over. Here is the initial expression for reference: 4 + 32 / 8 × (−2) + 20 / 5 / 2 − 8 Table A-4 is an S/R breakdown of the evaluation process. 10 in Chap. If you start with a positive integer p and multiply by −3. Evaluating this requires close attention and patience.Chapter 5
599
Table A-3. replace original parentheses with brackets for temporary clarification Get rid of brackets. you get a negative integer whose absolute value is 3 |p|. 3. and the absolute value increases by a factor of 3 every time.
Statements a + (b − c) + (a − b) + c a + [b + (−c)] + [a + (−b)] + c a + b + (−c) + a + (−b) + c a + a + b + (−b) + c + (−c) a+a Reasons Begin here Change subtractions into negative additions. they served a good purpose but are not necessary in straight sums Commutative law for addition. If you start with a negative integer n and multiply by −3. 2. forever As you can see. unless the starting integer q happens to be 0. whether positive or negative. you get a positive integer whose absolute value is 3 |n|. the polarity (that is. Solution to the second part of Prob. the absolute value grows rapidly as you keep on multiplying by −3. you get |−3q| = 3 |q| |−3 × (−3q)| = 3 × 3 |q| |−3 × (−3) × (−3q)| = 3 × 3 × 3 |q| ↓ and so on. each statement is equal to every statement above it. generalized b + (−b) = 0 and c + (−c) = 0 so the b’s and c’s “cancel out”
Chapter 5
1.
. the “positivity” or “negativity”) of the end product keeps alternating. So with any integer q. As you read down the left-hand column.

c. Here are the steps in the calculation process.
Statements 4 + 32 / 8 × (−2) + 20 / 5 / 2 − 8 4 + 32 / [8 × (−2)] + 20 / 5 / 2 − 8 4 + 32 / (−16) + 20 / 5 / 2 − 8 4 + [32/(−16)] + [(20/5)/2)] − 8 4 + (−2) + [(20/5)/2] − 8 4 + (−2) + (4/2) − 8 4 + (−2) + 2 − 8 4 + (−2) + 2 + (−8) −4 Reasons Begin here Group the multiplication Do the multiplication Group the divisions Do the division 32/(−16) = −2 Do the division 20/5 = 4 Do the division 4/2 = 2 Convert the subtraction to negative addition Do the additions from left to right
Table A-5. Solution to Prob. Then a can be any integer. and d. Follow each statement and reason closely so you’re sure how it follows from the statements before. 3 in Chap. abcd = dcba 5. b. See Table A-5. 5. As you read down the lefthand column.D.600 Worked-Out Solutions to Exercises: Chapters 1 to 9
Table A-4.
Statements abcd a(bcd ) a(dcb ) Reasons Begin here Group the last three integers Result of the “challenge” where it was proved that you can reverse the order of a product of three integers Commutative law for the product of a and (dcb) Ungroup the first three integers Mission accomplished
(dcb)a dcba Q.E. This always works if c = 1. using parentheses where needed: −15 × (−45) = 675 675 / (−25) = −27 −27 × (−9) = 243 243 / (−81) = −3 −3 × (−5) = 15 6. 5. and b can be any integer except 0. each statement is equal to every statement above it. As you read down the left-hand column. 4 in Chap. This proves that for any four integers a. Here’s the original equation: (a /b)/c = a /(b /c)
.
4. Solution to Prob. each statement is equal to every statement above it.

once for each of the three products above: (m + n)p = mp + np Q. again. always works if c = 1. can you guess what’s coming? Another extra-credit workout! Does the original rule hold true when a = 1.E. That will help keep you from sinking into an “abc rut” with the naming of variables. This proves that for any two integers d and g. −(h − k) = k − h
. Let’s start with the distributive law of multiplication over addition in its left-hand form.Chapter 5
601
“Plug in” 1 for c. Does the original rule work when a = 1. See Table A-7. Otherwise. This proves that for any two integers h and k. m. but you let a be any integer and c be any integer except 0? 8. Then you get (ab)/1 = a(b /1) which simplifies to ab = ab Now. and p are integers. the law is stated in exactly the same way. You’ll notice that we’re using different letters of the alphabet. but you let b be any integer and c be any integer except 0? Does it work when b = 1. but you let b and c be any integers except 0? Does it work when b = 1. unfamiliar variable names can keep your attention on the way things work. 8.D. the solution is only a matter of applying the commutative law for multiplication three times. We can write the original form of the law like this: p(m + n) = pm + pn where n. Then a and b can be any integers. The variables have unfamiliar names for the same reason as in Prob. This. Here’s the original equation: (ab)/c = a(b /c) “Plug in” 1 for c. but you let a be any integer and c be any integer except 0? 7. See Table A-6. without getting stuck in the “abc routine” of rote memorization. That’s all there is to it! 9. −(d + g) = −d − g 10. Again. Now. Then you get (a /b) /1 = a /(b/1) which simplifies to a /b = a /b Here is an opportunity get some extra credit.

The square root of 373 is equal to 19 followed by
. The ratio of the absolute temperature of the boiling point to the absolute temperature of the freezing point. 9 in Chap. expressed in kelvins. getting 31:19. 2.D. 1! Start by trying to factor 373 into a product of primes.602 Worked-Out Solutions to Exercises: Chapters 1 to 9
Table A-6. The ratio of the top wind speed in a “category 1” hurricane to the top wind speed in a “category 4” hurricane is 95:155. Solution to Prob. To get this in lowest terms.D. you get 19:31.
Statements −(d + g) (d + g)(−1) (−1)(d + g) (−1)d + (−1)g −d + (−g) −d − g Q. The ratio of the top wind speed in a “category 4” hurricane to the top wind speed in a “category 1” hurricane is found by switching the numerator and denominator. 10 in Chap. These integers are both prime. As you read down the left-hand column. That’s not as obvious here as in Prob. Solution to Prob. When you divide both the numerator and the denominator by 5. each statement is equal to every statement above it.E. Reasons Begin here Principle of the sign-changing element Commutative law for multiplication Distributive law for multiplication over addition Principle of the sign-changing element (the other way around) Addition of a negative is the same thing as subtraction Mission accomplished
Table A-7. each statement is equal to every statement above it. 5. 5. Reasons Begin here Principle of the sign-changing element Right-hand distributive law for multiplication over subtraction Principle of the sign-changing element (the other way around) Subtraction of a negative is the same thing as addition of a positive Commutative law for addition Addition of a negative is the same thing as subtraction of a positive Mission accomplished
Chapter 6
1.
Statements −(h − k) (h − k)(−1) h(−1) − k(−1) −h − (−k) −h + k k + (−h) k−h Q.E. is 373:273. you must find an integer that “divides out” from both of these numbers until the denominator becomes prime. so this expression of the ratio is in lowest terms. because they both end in 5. It’s easy to see that both of these integers are divisible by 5. As you read down the lefthand column.

consider this: 3/5 × 7/11 = (3 × 7) / (5 × 11) = 21/55 This product is in lowest terms. you’ll see that none of them gives you a quotient without a remainder. You can sense immediately that this is in lowest terms because the numerator and the denominator have absolute values that differ by only 1. Round this up to 20. Therefore. 7. But there’s an easier way to see this. the fraction 231/230 has to be in lowest terms. 13. You can use the “brute-force” method and factor both the numerator and the denominator of 231/230 into primes. so the ratio 373:273 is in lowest terms. You know this because the numerator is the product of the primes 3 and 7. and the denominator is positive. convert both the numerator and the denominator into products of primes and then attach the extra “factor” of −1 to the numerator. the product is sometimes in lowest terms. First. Now think: what will happen if you divide both the numerator and the denominator by any positive integer other than 1. and you don’t have to do any work to figure it out. and you’ll discover that the prime factors that make up the numerator are entirely different from the prime factors that make up the denominator. in an attempt to reduce the fraction? The resulting numerator and denominator will always have absolute values that differ by less than 1. but not always. 17. 3. When you have two fractions in lowest terms and multiply them. so they can’t both be integers. and the numerator has the extra “factor” −1: (−1 × 2 × 7 × 11) / (3 × 5 × 11) The common prime factor is 11. and 19. and the denominator is positive. the
. and the denominator is the product of the primes 5 and 11. Remove it from both the numerator and the denominator. When the numerator and denominator of a fraction are both factored into primes. 5. 3. to reduce it. getting (−1 × 2 × 7) / (3 × 5) That’s −14/15.” both the numerator and the denominator must be integers. 5.Chapter 6
603
a decimal point and some digits. 4. and if it is not. use these products to build a fraction in which both the numerator and the denominator consist of prime factors. like this: −154 = −1 × 2 × 7 × 11 and 165 = 3 × 5 × 11 Next. That means 373 is itself a prime number. Note that the absolute values of the numerator and the denominator differ by only 1. But in order to be a “legitimate fraction. 11. The primes less than 20 are 2. As you divide 373 by each of these primes. First. You want to see if −154/165 is in lowest terms.

but the product is not. and no prime in the numerator is the same as any prime in the denominator. this quotient is not in lowest terms.604 Worked-Out Solutions to Exercises: Chapters 1 to 9
denominator is positive. 6. You can factor 5 out of both the numerator and the denominator of 15/35. then that fraction is in lowest terms. reducing it to 3/7. as you saw in the solution to Prob. each statement is equal to all the statements above it. This can be seen by “mutating” the solution to Prob. where the second pair of fractions is divided first. Now try this: (3/5) / (7/5) = 3/5 × 5/7 = (3 × 5) / (5 × 7) = 15/35 Again as you saw in the solution to Prob.
Statements (a /b) / [(c /d ) / (e /f )] (a /b) / (cf / de) (a /b) / (g /h) ah / bg ade / bcf Reasons Begin here Apply the formula for division of c /d by e /f Temporarily let cf = g and de = h. it’s not hard to think of a couple of fractions that are in lowest terms individually. With this knowledge. 7. consider this: (3/5) / (11/7) = 3/5 × 7/11 = (3 × 7) / (5 × 11) = 21/55 This is in lowest terms. Table A-8 is an S/R derivation proving that (a /b) / [(c /d ) / (e /f )] = ade / bcf
Table A-8. 6. Solution to Prob. and substitute the new names in the previous expression Apply the formula for division of a /b by g /h Substitute cf for g and de for h in the previous expression
. First. 5. 5. 7 in Chap. 5. even though the dividend and divisor fractions are. As you read down the left-hand column. but not always. but that produce a fraction that is not in lowest terms when they are multiplied by each other. the quotient is sometimes in lowest terms. Take a look at this: 3/5 × 5/7 = (3 × 5) / (5 × 7) = 15/35 Both of the original fractions are in lowest terms here. The result is a formula for repeated division of fractions. When you have two fractions in lowest terms and divide one by the other.

Chapter 6
605
Table A-9.
. a=a The symmetric property tells us that for any two quantities a and b. then b = a The transitive property tells us that for any three quantities a. then a = c Whenever any means of comparing things has all three of these properties. often with a little extra space on either side (⇔). then it’s called an equivalence relation. and d are nonzero integers. and transitive properties. where you found out that [(a /b) / (c /d )] / (e /f ) = adf / bce 8.D. Reasons Begin here Formula for multiplication of fractions Commutative law for multiplication of integers applied to numerator and denominator Formula for multiplication of fractions applied “backwards” Mission accomplished
Compare this with the result of the “challenge” problem. 6. Table A-9 is an S/R proof that if a. each statement is equal to all the statements above it. This shows that the commutative law holds for the multiplication of fractions. and c.E. 8 in Chap. If a = b and b = c. then (a /b)(c /d ) = (c /d )(a /b) Therefore. in which the rule for multiplication of fractions is applied “backwards. b. the commutative property holds for the multiplication of fractions. b.” symbolized by a double-shafted. which may seem trivial to you but is really quite significant. You should know what these terms mean. such as the logical connector “if and only if ” or “iff. double-headed arrow. is one of three aspects of equality known as the reflexive. c. If a = b. Equality is the most common example of an equivalence relation. But there are others.” We can get away with this because equality works in both directions! This fact. symmetric. and indeed for the multiplication of any two rational numbers. Solution to Prob. The reflexive property tells us that for any quantity a. Note the fourth step of this proof. As you read down the left-hand column.
Statements (a /b)(c /d ) ac / bd ca / db (c /d )(a /b) Q.

This shows that the associative law holds for the multiplication of fractions.” into the form we want. in two “steps back down. 10. 9 in Chap.E. and a third of the form e /f. Then we invoke that law. and b. and d. and f are positive integers. We’ve been told that there are four integers a. Table A-10 shows that the associative property of multiplication holds for any three fractions.
9. Finally we mold the expression. c. If we have fraction of the form a /b. then [(a /b)(c /d )](e /f ) = (a /b)[(c /d )(e /f )] Do you notice that this proof has a certain symmetry? First. in two “steps up. d.
Statements [(a /b)(c /d )](e /f ) (ac / bd )(e /f ) (ac)e / (bd )f a(ce) / b(df ) (a /b)(ce / df ) Reasons Begin here Formula for multiplication of a /b by c /d Formula for multiplication of (ac / bd ) by e /f Associative law for multiplication of integers applied to numerator and denominator Formula for multiplication of fractions applied “backward” to turn quotient of products into product of quotients Formula for multiplication of fractions applied “backward” (again!) to turn quotient of products into product of quotients Mission accomplished
(a /b )[(c /d )(e /f )]
Q. we get ad / bc = cb / da Invoking the commutative law for multiplication in the numerator and denominator on the right-hand side of the equals sign.” into a form where we can apply the associative law for multiplying integers. each statement is equal to all the statements above it. we see that ad / bc = bc / ad
. such that (a /b) / (c /d ) = (c /d ) / (a /b) If we use the rule for division of one fraction by another on both sides of this equation. and e are integers. As you read down the left-hand column.D. 6. Solution to Prob. where a.606 Worked-Out Solutions to Exercises: Chapters 1 to 9
Table A-10. we mold the expression. Let’s explore this situation and see what sorts of “clues” we can find. b. another of the form c /d. c.

Chapter 7

607

We aren’t allowed to get away with trivial solutions, such as letting all the integers be equal to 1 or letting them all be equal to −1. But suppose that a = 7, b = 5, c = 14, and d = 10. (This isn’t the only example we can use, but it should give you the general idea.) Then ad / bc = (7 × 10) / (5 × 14) = 70/70 =1 and bc / ad = (5 × 14) / (7 × 10) = 70/70 =1 Let’s “plug in” the values a = 7, b = 5, c = 14, and d = 10 to the original equation and see what we get: (a /b) / (c /d) = (c /d) / (a /b) therefore (7/5) / (14/10) = (14/10) / (7/5) Note that 7/5 and 14/10 actually represent the same rational number. All we’ve really done here is show that 1 = 1, in a roundabout way.

Chapter 7
1. Figure A-1 shows a number line that covers the range of positive rational numbers from 10 up to 100,000. To find the number of orders of magnitude, subtract the powers of 10. Figure A-1 Illustration for the solution
to Prob. 1 in Chap. 7.
5

10

100,000

10

4

10,000

10

3

1,000

10

2

100

10

1

10

608 Worked-Out Solutions to Exercises: Chapters 1 to 9

Figure A-2 Illustration for the solution
to Prob. 2 in Chap. 7.

3 × 10 3 × 10 3 × 10 3 × 10

5

300,000

4

30,000

3

3,000

2

300

3 × 10

1

30

Here, 100,000 = 105 and 10 = 101. We have 5 − 1, or 4, so there are four orders of magnitude in this span. Another way to see this is to count the number of intervals between “hash marks” on the number line. 2. Figure A-2 shows a number line that covers the range of positive rational numbers from 30 up to 300,000. In this case, we can divide the larger number by the smaller, and then count the number of ciphers in the quotient. We get 300,000 / 30 = 10,000 These two numbers differ by four orders of magnitude, the same extent as the two numbers differ in Prob. 1. We can also count the number of intervals between hash marks, just as in Prob. 1. 3. In this situation, the larger number is 75,000,000 and the smaller number is 330. If we divide the larger by the smaller, we get 227,272 and a fraction. That’s between a factor of 100,000 (or 105) and 1,000,000 (or 106). We can therefore say that 75,000,000 is between five and six orders of magnitude larger than 330. Right now, that’s the best we can do. (It’s possible to come up with a more precise value, but that involves logarithms, which we have not yet studied.) 4. The answers, along with explanations, are as follows: (a) The number 4.7 is equivalent to 4-7/10. The fractional part here is in lowest terms, because the numerator, 7, is prime. (b) The number −8.35 is equivalent to −8-35/100. The fractional part can be reduced to 7/20, so the entire expression becomes −8-7/20. Note the difference in appearance, as well as the difference in purpose, between the minus sign and the dash! (c) The number 0.02 has no integer portion. The fractional part is 2/100, which reduces to 1/50. Therefore, the entire expression is 1/50. (d) The number −0.29 has no integer portion. The fractional part is −29/100, which is already in lowest terms because the absolute value of the numerator, 29, is prime and is not a prime factor of the denominator. Therefore, the entire expression is −29/100.

Chapter 7

609

5. The answers, along with explanations, are as follows: (a) We have the number 4-7/10. We can convert the integer part, which is 4, into 10ths by multiplying it by 10 and then dividing the result by 10, getting (4 × 10) / 10 = 40/10 The entire number is therefore 40/10 + 7/10 = (40 + 7) / 10 = 47/10 (b) We start with the number −8-7/20. We convert the integer part, −8, into 20ths by multiplying and then dividing by 20, getting (−8 × 20) / 20 = −160/20 The entire number is therefore −160/20 + (−7/20) = (−160 − 7) / 20 = −167/20 (c) The number 1/50 is a fraction already, and has been reduced to lowest terms. There’s nothing further for us to do here! (d) This situation is the same as in part (c). We already have the final expression in the form of the starting number, −29/100. 6. The easiest way to work out these problems is to input the numerator into a calculator, and then divide by the denominator. When we do that, we get the following results. (a) 44/16 = 2.75 (b) −81/27 = −3 (c) 51/13 = 3.923076923076923076... (d) −45/800 = −0.05625 In case (c), you’ll need a calculator that can display a lot of digits if you want to be certain of the repeating pattern of digits to the right of the decimal point. (It’s 923076.) If you don’t have such a calculator, you can perform old-fashioned, manual long division to discover the pattern. 7. This problem can be solved in two steps. First, we use a calculator or long division to determine the decimal equivalent of 1/17. We get this endless repeating decimal: 0.05882352941176470588235294117647... The repeating sequence of digits is 0588235294117647. The initial cipher is important here! Next, we count up the number of digits in this sequence, including the cipher at the beginning. There are 16 digits in the repeating string. We construct a fraction with the

610 Worked-Out Solutions to Exercises: Chapters 1 to 9

repeating sequence in the numerator and a string of 16 digits, all 9s, in the denominator, inserting commas to make the large numbers more readable. That gives us 0,588,235,294,117,647 / 9,999,999,999,999,999 The initial cipher can now be removed. It was only necessary to be sure we put the correct number of 9s in the denominator. The final answer is: 588,235,294,117,647 / 9,999,999,999,999,999 If you want, you can check this by dividing it out using a calculator with a large display, such as the one in a computer. You should get the same result as we got when we divided 1 by 17. Another way to check this is to divide the denominator of the above fraction by the numerator (that is, take the quotient “upside-down”). You should get exactly 17. 8. The number we are given to start with is 2.892892892.... To convert this to a ratio of integers, we first write down the part of the expression to the right of the decimal point, like this: . 892 892 892 ... From this, we know that fractional part of the expression is 892/999. We put back the whole-number part, getting 2-892/999 Now we must convert 2 to a fraction with a denominator 999. We multiply 2 by 999, getting 1,998. This goes into the numerator. Now we have two fractions that we can easily add to produce the final ratio: 892 / 999 + 1,998 / 999 = (892 + 1,998) / 999 = 2,890 / 999 It is always a good idea to check the results of calculations like this by dividing out on a calculator. In this case, the quotient is 2.892892892..., the original number in decimal form. 9. We can be certain that this decimal expansion, which we have been told is an endless string of digits, has a repeating pattern. The original quotient is a rational number by definition. Remember, any rational number can be expressed as either a terminating decimal or an endless repeating decimal. The repeating pattern of digits in the decimal expansion might be incredibly long, but it is finite. 10. This is one of the most baffling problems in mathematics. The trouble comes up because we’re trying to compare hard reality with pure theory. Even the most powerful supercomputer can be confused by a string-of-digits problem if the repeating pattern is complicated enough. But the fact that a pattern can’t be discovered in a human lifetime does not prove conclusively that there is not a pattern! It works the other way, too. If we see a long string of digits repeating many times, we can’t be sure it will repeat endlessly, unless we know that there’s a ratio of integers with the same value.

Chapter 8

611

Chapter 8
1. When a negative number is raised to an even positive integer power, the result is always a positive number. When a negative number is raised to an odd positive integer power, the result is always a negative number. 2. The answers, along with explanations, are as follows. (a) If we raise a base of −2 to increasing integer powers starting with 1, we get this sequence: (−2)1, (−2)2, (−2)3, (−2)4, (−2)5, ... When we multiply these out, we get −2, 4, −8, 16, −32, ... The numbers alternate between negative and positive, and their absolute values double with each repetition. This sequence “runs away” toward both “positive infinity” and “negative infinity”! (b) If we do the same thing with a base of −1, we get (−1)1, (−1)2, (−1)3, (−1)4, (−1)5, ... Multiplying these out gives us −1, 1, −1, 1, −1, ... The numbers simply alternate between −1 and 1. (c) If we carry out the same process with a base of −1/2, we get (−1/2)1, (−1/2)2, (−1/2)3, (−1/2)4, (−1/2)5, ... Multiplying these out produces −1/2, 1/4, −1/8, 1/16, −1/32, ... The numbers again alternate between negative and positive, and their absolute values get half as large with each repetition. This sequence converges toward 0 “from both sides.” 3. Here are the answers. Note how they “mirror” the results of Prob. 2. (a) If we raise a base of −2 to smaller and smaller negative integer powers starting with −1, we get this sequence: (−2)−1, (−2)−2, (−2)−3, (−2)−4, (−2)−5, ... This is the same as 1/(−2)1, 1/(−2)2, 1/(−2)3, 1/(−2)4, 1/(−2)5, ...

612 Worked-Out Solutions to Exercises: Chapters 1 to 9

When we multiply these out, we get 1/(−2), 1/4, 1/(−8), 1/16, 1/(−32), ... which is the same as −1/2, 1/4, −1/8, 1/16, −1/32, ... The numbers alternate between negative and positive, and their absolute values get half as large with each repetition. This sequence is identical to the solution for Problem 2(c). (b) If we do the same thing with a base of −1, we get (−1)−1, (−1)−2, (−1)−3, (−1)−4, (−1)−5, ... This is the same as 1/(−1)1, 1/(−1)2, 1/(−1)3, 1/(−1)4, 1/(−1)5, ... Multiplying these out gives us 1/(−1), 1/1, 1/(−1), 1/1, 1/(−1), ... which is the same as −1, 1, −1, 1, −1, ... The numbers simply alternate between −1 and 1. This result is identical with the solution for Prob. 2(b). (c) Finally, let’s do the process with a base of −1/2. We get the sequence (−1/2)−1, (−1/2)−2, (−1/2)−3, (−1/2)−4, (−1/2)−5, ... This is the same as 1/(−1/2)1, 1/(−1/2)2, 1/(−1/2)3, 1/(−1/2)4, 1/(−1/2)5, ... which is the same as 1/(−1/2), 1/(1/4), 1/(−1/8), 1/(1/16), 1/(−1/32), ... which can be simplified to −2, 4, −8, 16, −32, ... That’s the same thing we got when we solved Problem 2(a).

Chapter 8

613

4. The expression can be simplified to a sum of individual terms when we remember that squaring any quantity (that is, taking it to the second power) is the same thing as multiplying it by itself. Then we can use the results of the final “Challenge” in Chap. 5. Step-by-step, we get: ( y + 1)2 = ( y + 1)( y + 1) = yy + y1 + 1y + (1 × 1) = y2 + y + y + 1 = y 2 + 2y + 1 5. This problem can be solved just like Prob. 4, but we have to pay careful attention because of the minus sign: ( y − 1)2 = ( y − 1)( y − 1) = yy + y(−1) + (−1y) + [(−1) × (−1)] = y 2 + (−y) + (−y) + 1 = y 2 − 2y + 1 6. Let’s start with the generalized addition-of-exponents (GAOE) rule as it is stated in the chapter text. Here it is again, with the exponent names changed for variety! For any number a except 0, and for any rational numbers p and q, a pa q = a ( p + q) Suppose r is another rational number. Let’s multiply both sides of the above equation by a r. This gives us (a pa q)a r = a( p + q)a r According to the rule for the grouping of factors in a product, we can take the parentheses out of the left-hand side of the above equation and get a pa qa r = a( p + q)a r The left-hand side of the equation now contains the expression we want to evaluate. Let’s consider (p + q) to be a single quantity. We can then use the GAOE rule on the righthand side of this equation, getting a pa qa r = a[( p + q) +r] Again, we invoke the privilege of ungrouping, this time to the addends in the exponent on the right-hand side. This gives us a pa qa r = a( p + q + r) Q.E.D. Mission accomplished!

614 Worked-Out Solutions to Exercises: Chapters 1 to 9

Table A-11. Solution to Prob. 7 in Chap. 8. This shows that the multiplication-of-exponents rule applies to a “power of a power of a power.” As you read down the left-hand column, each statement is equal to all the statements above it.
Statements (a p)q = a pq [(a p)q]r = (a pq)r [(a p)q]r = a (pq)r [(a p)q]r = a pqr Q.E.D. Reasons GMOE rule as given in Chap. 8 text, where a is any number except 0, and p and q are rational numbers Take r th power of both sides, where r is a rational number Consider (pq) as a single quantity and then use GMOE rule on right-hand side Ungrouping of products in exponent on right-hand side Mission accomplished

7. This S/R proof is shown in Table A-11. We can “legally” take the r th power of both sides in line 2 of the proof even if r is a reciprocal power. Remember, if there is any positivenegative ambiguity when taking a reciprocal power, the positive value is the “default.” 8. We can start by stating the generalized multiplication-of-exponents (GMOE) rule as it appears in the chapter text. The exponent names are changed to keep us out of a “rote-memorization rut,” and also to conform to the way the problem is stated. For any number x except 0, and for any rational numbers r and s, (x r )s = x rs Applying the commutative law for multiplication to the entire exponent on the righthand side of this equation, we get (x r )s = x sr Finally, we can invoke the GMOE rule “in reverse” to the right-hand side, obtaining (x r )s = (x s )r Q.E.D. Mission accomplished! 9. Let’s suppose that x is a positive number. It can by any number we want, as long as it is larger than 0. We take the 4th root or 1/4 power of x, and then square the result. That gives us (x 1/4)2 According to the GMOE rule, that is the same as x (1/4) × 2

Chapter 9

615

which simplifies to x 2/4. The fraction 2/4 can be reduced to 1/2. That means we actually have x raised to the 1/2 power, or the square root of x. 10. Imagine that y is a positive number. We take the 6th power of y, and then take the cube root or 1/3 power of the result. That gives us (y 6)1/3 According to the GMOE rule, that is the same as y 6 × (1/3) which simplifies to y 6/3 and then reduces to y 2.

Chapter 9
1. We can suspect that quantity (d), 271/2, is irrational. It is not a natural number. When we use a calculator to evaluate it, we get 5.196... followed by an apparently random jumble of digits. That suggests its decimal expansion is endless and nonrepeating. The other three quantities can be evaluated and found rational: (a) 163/4 = (161/4)3 = 23 = 8 (b) (1/4)1/2 = 1/(41/2) = 1/2 (c) (−27)−1/3 = 1/(−271/3) = 1/(−3) = −1/3 2. If we have an irrational number expanded into endless, nonrepeating decimal form, we can multiply it by any natural-number power of 10 and always get the same string of digits. The only difference will be that the decimal point moves to the right by one place for each power of 10. As an example, consider the square root of 7, or 71/2. Using a calculator with a large display, we see that this expands to 71/2 = 2.64575131106459059... As we multiply by increasing natural-number powers of 10, we get this sequence of numbers, each one 10 times as large as the one above it: 10 × 71/2 = 26.4575131106459059... 100 × 71/2 = 264.575131106459059... 1,000 × 71/2 = 2,645.75131106459059... 10,000 × 71/2 = 26,457.5131106459059... ↓ and so on, as long as we want These are all endless non-repeating decimals, so they’re all irrational numbers. This will happen for any endless non-repeating string of digits.

616 Worked-Out Solutions to Exercises: Chapters 1 to 9

3. All of these sets are infinite, and the elements of each set can be completely defined by means of an “implied list.” Therefore, the cardinality of every one of these sets is ℵ0. We can pair off any of these sets one-to-one with the set N of naturals. As an optional exercise, you might want to show how this can be done. Here’s a hint: Multiply the naturals all by 2, 10, 100, or any whole-number power of 10 to create “implied lists” for the sets. 4. The original equation is 36x + 48y = 216 When we divide through by 12 on either side, we get (36x + 48y)/12 = 216/12 which can be morphed to (36/12)x + (48/12)y = 18 and finally to 3x + 4y = 18 That’s as simple as we can get it. 5. To figure this out, note that 18 = 9 × 2. Therefore, according to the power of product rule, we have 181/2 = (9 × 2)1/2 = 91/2 × 21/2 = 3 × 21/2 This is a product of a natural number and an irrational number. The nonnegative square roots of large numbers can often be resolved in this way. 6. The number 83 is prime. If we want to factor this into a product of two natural numbers and no remainder, the best we can do is 83 × 1. Therefore, we can’t resolve the nonnegative square root of 83 into anything simpler than 831/2. We can also tell that it is in the most simplified form because it has no factors that are perfect squares. 7. The ratio of 501/2 to 21/2 is the same thing as the quotient of these two numbers. Note that both 50 and 2 are taken to the same real-number power, that is, 1/2. According to the power of quotient rule, then, we have 501/2/21/2 = (50/2)1/2 = 251/2 =5

Chapter 9

617

The ratio of the nonnegative square root of 50 to the nonnegative square root of 2 is exactly equal to 5, even though the dividend and the divisor are both irrational. 8. Here’s the sum-of-quotients rule again: w /x + y /z = (wz + xy)/(xz) In this example, we can let w = 7, x = 11, y = −5, and z = 17. None of these are equal to 0, so we can be sure the rule will work properly. Now it’s simply a matter of doing the arithmetic: 7/11 + (−5/17) = {(7 × 17) + [11 × (−5)]} / (11 × 17) = [119 + (−55)] / 187 = 64 / 187 This can’t be reduced to lower terms because the denominator, 187, is the product of two primes, 11 and 17. Neither of these factors “goes into” the numerator, 64, without leaving a remainder. 9. To solve this, we can rewrite (x − y) as [x + (−y)]. Then, using the product of sums rule, we have (x + y)[x + (−y)] = xx + x (−y) + yx + y (−y) We can use familiar arithmetic rules to write this as = x 2 − xy + xy − y 2 The addends −xy and xy cancel out here, so we get the final result = x2 − y2 10. Let u, v, w, x, y, and z be real numbers. We’re given a product of sums of three variables, and we’re told to multiply it out using the product of sums rule. That rule, as stated in the chapter text, only allows us to use two variables in each addend. But we can “cheat” by renaming certain sums! Here is the original expression: (u + v + w)(x + y + z) Let’s rename (v + w) as r, and (y + z) as s. Then we have (u + r)(x + s) Using the product of sums rule, we can multiply this out to ux + us + rx + rs Now let’s substitute the original values for r and s back into the expression. This gives us ux + u(y + z) + (v + w)x + (v + w)(y + z)

618 Worked-Out Solutions to Exercises: Chapters 1 to 9

The first addend in this expression is ready to go! We can use the distributive laws on the second and third terms to get ux + uy + uz + vx + wx + (v + w)(y + z) The final term can be multiplied out using the product of sums rule from the chapter text. This gives us ux + uy + uz + vx + wx + vy + vz + wy + wz We can use the commutative law of addition, in its generalized form, to rearrange these terms so the whole sum is easier to remember: ux + uy + uz + vx + vy + vz + wx + wy + wz

APPENDIX

B

Worked-Out Solutions to Exercises: Chapters 11 to 19
These worked-out solutions do not necessarily represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see here, by all means go ahead! But always check your work to be sure your “alternative” answer is correct.

or −7 = −7 = −7 3. If we multiply an equation through by the number 0, we will always get a statement to the effect that 0 equals itself. That’s true, but it’s trivial and isn’t good for much of anything. We might multiply an equation through by a variable or expression that’s equal to 0, even though we aren’t aware of it at the time. That’s likely to make the equation more complicated, but it won’t make it false. Consider this: x=2 Let’s multiply this through by (2 − x). We get x (2 − x) = 2(2 − x) which expands to 2x − x 2 = 4 − 2x In this case, our manipulation does us no good. But it does no real harm either, as the inadvertent division by 0 can do. Occasionally, a manipulation like this can put a complicated equation into a form that’s easier to work with. 4. Let’s call the set of negative integers Z−. Remember the standard name for the set of natural numbers; it’s N. We have Z− = {..., −5, −4, −3, −2, −1} and N = {0, 1, 2, 3, 4, 5, ...} From these statements, we can see that any negative integer we choose will be smaller than any natural number we choose. Therefore, if x is an element of Z− and y is an element of N, x is smaller than y. In logical form along with set notation, we can write this as [(x ∈ Z−) & (y ∈ N )] ⇒ x < y 5. Let’s call the set of nonpositive reals R0− (for “0 and all the negative reals”) and the set of nonnegative reals R0+ (for “0 and all the positive reals”). Both of these sets include 0, but that’s the only element they share. Therefore, any nonpositive real number we choose must be smaller than or equal to any nonnegative real number we choose. In other words, if x is an element of R0− and y is an element of R0+, then x is smaller than or equal to y. In logical form along with set notation, we can write this as [(x ∈ R0−) & ( y ∈ R0+)] ⇒ x ≤ y

Chapter 11

621

6. We have standard names for the sets of rational and irrational numbers: Q and S, respectively. These sets are disjoint; they have no elements in common. If x is an element of Q and y is an element of S, then x is never equal to y. In logical form along with set notation, we can write [(x ∈ Q) & (y ∈ S )] ⇒ x ≠ y 7. We can write the statement as “mathematical verse” by reading it out loud and taking careful note of each symbol. Here’s the logical statement again, for reference. (∀ a, b, c) : [(a ≥ b) & (b ≤ c)] ⇒ (a = c) When we break up the statement into parts and write them down on separate lines, we come up with the following: For all a, b, and c : If a is larger than or equal to b, and b is smaller than or equal to c, then a is equal to c. This little poem might be cute, but it doesn’t state a valid mathematical law. Suppose that a = 5, b = 3, and c = 7. In that case, a is larger than or equal to b and b is smaller than or equal to c. However, a is not equal to c. 8. Our task is to simplify the equation to a form where x appears all by itself on the left side of the equality symbol, and a plain numeral appears all by itself on the right. Here’s the equation again, for reference: x + 4 = 2x We can subtract x from both sides, getting x + 4 − x = 2x − x which simplifies to 4=x We can reverse the order to get the solution in its proper form: x=4 The original equation holds true only when x is equal to 4.

622 Worked-Out Solutions to Exercises: Chapters 11 to 19

9. We must simplify the inequality so that y appears all by itself on the left side of the “not equal” symbol, and a plain numeral appears all by itself on the right. Here’s the inequality again, for reference: y /2 ≠ 4y + 7 First, let’s multiply through by 2. That gives us (y /2) × 2 ≠ (4y + 7) × 2 which multiplies out to y ≠ 8y + 14 Now, let’s subtract 8y from each side. That gives us y − 8y ≠ 8y + 14 − 8y which simplifies to −7y ≠ 14 We can divide this through by −7 to get (−7y)/(−7) ≠ 14/(−7) which simplifies to y ≠ −2 The original inequality holds true for all values of y except −2. 10. We must simplify the inequality so that z appears all by itself on the left side of the “smaller than or equal” symbol, and a plain numeral appears all by itself on the right. Here’s the inequality again, for reference: z /(−3) ≤ 6z + 6 Let’s multiply through by −3, remembering that we must reverse the sense of the inequality whenever we multiply through by a negative. That gives us [z /(−3)] × (−3) ≥ (6z + 6) × (−3) which simplifies to z ≥ −18z − 18

Chapter 12

623

If we add 18z to each side, we get z + 18z ≥ −18z − 18 + 18z which simplifies to 19z ≥ −18 We finish by dividing each side by 19. That leaves us with z ≥ −18/19 The original inequality holds true for all values of z larger than or equal to −18/19.

Chapter 12
1. See Table B-1. 2. See Table B-2. 3. See Table B-3.

Table B-1.
Statements 4x + 4 = 2x − 2 2x + 4 = −2 2x + 6 = 0

Solution to Prob. 1 in Chap. 12.
Reasons This is the equation we are given Subtract 2x from each side Add 2 to each side

Solution to Prob. 2 in Chap. 12.
Reasons This is the equation we are given Multiply through by 3 Distributive law applied to the right side Subtract 18x from each side Subtract 6 from each side Multiply through by −1 and apply the distributive law to the right side, obtaining a more elegant equation

Solution to Prob. 3 in Chap. 12.
Reasons This is the equation we are given Multiply through by 7 Apply distributive law to each side Simplify the right side Subtract 50x from each side Multiply through by −1 and apply the distributive law to the right side, obtaining a more elegant equation

4. We are given the equation x /3 + x /6 = 12 Multiplying through by 6, we get 6(x /3 + x /6) = 72 When we apply the distributive law to the left side and then simplify the addends, we obtain 2x + x = 72 Simplifying the left side further, we get 3x = 72 Subtracting 72 from each side yields 3x − 72 = 0 Finally, we can divide through by 3 to obtain lowest terms: x − 24 = 0 5. When we have a first-degree equation in standard form, the solution process never takes more than two steps. Let’s solve the results of Probs. 1 through 4 For Prob. 1 (Table B-1). The standard-form equation we got was 2x + 6 = 0 We subtract 6 from each side, getting 2x = −6

Chapter 12

625

Then we divide through by 2, getting x = −3 For Prob. 2 (Table B-2). The standard-form equation we got was 17x + 6 = 0 We subtract 6 from each side, getting 17x = −6 Then we divide through by 17, getting x = −6/17 For Prob. 3 (Table B-3). The standard-form equation we got was 43x + 49 = 0 We subtract 49 from each side, getting 43x = −49 Then we divide through by 43, getting x = −49/43 For Prob. 4. The standard-form equation we got was x − 24 = 0 We add 24 to each side, getting x = 24 In this case, we need not multiply or divide through by anything, because x is already by itself on the left side of the equation. 6. If we call the unknown number x, then we have the following equation to solve: (2x + 8)/4 = −1

626 Worked-Out Solutions to Exercises: Chapters 11 to 19

Multiplying through by 4, we get 2x + 8 = −4 Subtracting 8 from each side gives us 2x = −12 Dividing through by 2 produces x = −6 7. If we call the unknown number x, then we have the following equation to solve: (x − x /10)/2 = 135 Multiplying through by 2 gives us x − x /10 = 270 Note that x − x /10 is the same as x − (1/10)x, or (9/10)x. So we have (9/10)x = 270 Multiplying through by 10/9 gives us (10/9)(9/10)x = 10/9 × 270 which simplifies to x = 300 8. Let’s call Bonnie’s weight, in kilograms (kg), the unknown x. Then Bruce’s weight in kilograms is x + 5, and Bill’s weight in kilograms is (x + 5) + 10, or x + 15. The total weight is 200 kg. We can now write the equation x + (x + 5) + (x + 15) = 200 Applying the commutative law and adding the constants on the left side gives us x + x + x + 20 = 200 which simplifies to 3x + 20 = 200

Chapter 13

627

Subtracting 20 from each side gives us 3x = 180 Dividing through by 3, we get x = 60 That means Bonnie weighs 60 kg. Bruce weighs 5 kg more than Bonnie, or 65 kg. Bill weighs 10 kg more than Bruce, or 75 kg. 9. We can work through this without resorting to equations. It’s first-degree algebra in a single variable, but we don’t have to express it symbolically! Let’s start by figuring out the land speed of the boat as it traveled upstream. The distance from our cabin to our cousin’s cabin is 18 miles (mi), and it took 1 h 12 min for the boat to travel that far. Because 12 min = 1/5 h, the trip took 6/5 h. The speed of the boat relative to the land was therefore 18 mi per 6/5 h, or 18/(6/5) = 18 × 5/6 = 15 mi/h If there were no current, the boat would have traveled at 18 mi/h relative to the water and relative to the land. The water speed was indeed 18 mi/h, but the land speed was only 15 mi/h. The current slowed our boat down by 18 − 15, or 3 mi/h. The river must therefore have been flowing at 3 mi/h. 10. When we go downstream, the river current will add 3 mi/h to our boat’s land speed. Again, if there were no current, the land and water speeds would both be 18 mi/h. That means the downstream speed of the boat will be 18 + 3, or 21 mi/h. We must travel a land distance of 18 mi. Time equals distance divided by speed. Therefore, if we let t represent the time in hours, we have t = 18/21 = 6/7 h Because 1 h = 60 min, 6/7 hour = 6/7 × 60 min, or, approximately, 51.43 min.

Chapter 13
1. For every boy there is exactly one girl dance partner and vice-versa, so the mapping is an injection. It’s also onto the entire set of girls. No girl has to sit out the dance without a partner, so it’s a surjection. Any mapping that’s both an injection and a surjection is, by definition, a bijection. 2. This mapping is not one-to-one, so it’s not an injection. That means it can’t be a bijection. However, it’s a surjection, because it’s onto the entire set of girls.

628 Worked-Out Solutions to Exercises: Chapters 11 to 19

3. This mapping, like the one in the Prob. 2, is not one-to-one, so it can’t be an injection or a bijection. But it’s a surjection, because it’s onto the entire set of boys. 4. Our mapping was not an injection, because it was not one-to-one. If it wasn’t an injection, it couldn’t have been a bijection. It was a surjection from A to B, because we mapped our message onto the entire set B, the set of all 175,000 subscribers to Internet Network Beta. The maximal domain was A, the set of all 60,000 subscribers to Internet Network Alpha (before we were kicked out). The essential domain was the set containing only you and me. The co-domain and the range were both the whole set B. 5. This relation is one-to-one. For every integer q, there is exactly one even integer r, which we can get by doubling q. Conversely, for every even integer r, there is exactly one integer q, which we get when we divide y by 2. Our relation is therefore an injection from Q to R. It is not onto R, however, because there are plenty of real numbers that are not even integers. That means we do not have a surjection from Q onto R. If it’s not a surjection, then it can’t be a bijection. 6. This relation is not one-to-one. For example, q = 3/7 and q = 3/11 both map into z = 3. That means it’s not an injection, so it cannot be a bijection, either. It is a surjection, because it’s onto the entire set Z of integers. We can choose any integer z, place it into the numerator of a fraction, choose a denominator such that the fraction is in lowest terms, and have a rational number q that maps to z. 7. This relation is not quite one-to-one. For every nonzero integer z, there is exactly one rational number q = 1/z. But if z = 0, there is no q such that q = 1/z. The relation is therefore not injective nor bijective. It is not surjective, either. There are plenty of rational numbers that are not reciprocals of integers; 3/7 is an example. 8. This relation is one-to-one. We “patched the hole” in the relation defined in Prob. 7. We declared that if z = 0, then q = 0. There is no nonzero integer z such that 1/z = 0, so we don’t get a “dupe” by making this declaration. We have an injective relation now! But as with the relation in Prob. 7, we do not have a surjection. That means the relation is not bijective. 9. This relation is a function. For any value of x we choose, there is exactly one value of y such that y = x4. It’s not one-to-one between the set of reals and the set of nonnegative reals, but two-to-one except when x = 0. That means the function is not injective, so it can’t be bijective, either. It’s surjective, because it’s onto the entire set of nonnegative reals. For any nonnegative real number y, we can find a real number x such that y = x4. 10. When we transpose the values of the independent and dependent variables while leaving their names the same, we get x = y4 which is equivalent to y = ±(x1/4) The plus-or-minus symbol indicates that for every nonzero x, there are two values of y, one positive and the other negative. This relation is one-to-two except when x = 0, so it

Chapter 14

629

is not a function. It is not an injection because it’s not one-to-one. That means it cannot be a bijection. The relation does map onto its entire range (the set of all reals), so it’s a surjection.

Chapter 14
1. If we multiply x by −1 and leave y the same, the point will move to the other side of the y axis, but it will stay on the same side of the x axis. If it starts out in the first quadrant, it will move to the second. If it starts out in the second quadrant, it will move to the first. If it starts out in the third quadrant, it will move to the fourth. If it starts out in the fourth quadrant, it will move to the third. The y axis will act as a “point reflector.” 2. If we multiply y by −1 and leave x the same, the point will move to the other side of the x axis, but it will stay on the same side of the y axis. If it starts out in the first quadrant, it will move to the fourth. If it starts out in the second quadrant, it will move to the third. If it starts out in the third quadrant, it will move to the second. If it starts out in the fourth quadrant, it will move to the first. The x axis will act as a “point reflector.” 3. The point for (6x, 6y) will be in the same quadrant as the point for (x, y), but 6 times as far from the origin. The point for (x/4, y/4) will be in the same quadrant as the point for (x, y), but 1/4 of the distance from the origin. The origin, the point for (x, y), the point for (6x, 6y), and the point for (x/4, y/4) will all lie along a single straight line. 4. If the vertical test line intersects the graph (once for a function, and once or more for a relation), then the point where the test line intersects the independent-variable (horizontal) axis represents a numerical value in the domain. If the test line does not intersect the graph, then the point where the test line intersects the horizontal axis does not represent a value in the domain. 5. This process works just like the process for determining whether or not a point is in the domain, except that everything is rotated by 90°! If the horizontal test line intersects the graph, then the point where the test line intersects the dependent-variable (vertical) axis represents a numerical value in the range. If the test line does not intersect the graph, then the point where the test line intersects the vertical axis does not represent a value in the range. 6. We can plot several specific points for y = |x |, and then we can determine the graph on the basis of those points. We can deduce, using some common sense, that the lines are straight. See Fig. B-1. The vertical-line test tells us that this is a function of x. 7. We can plot several specific points for y = |x + 1|, and then we can determine the graph on that basis. Again, we can deduce, using some common sense, that the lines are straight. See Fig. B-2. The vertical-line test reveals that this is a function of x. 8. Figure B-3 is a graph of the inverse of y = x + 1. If we apply the “point reflector” method to Fig. 14-11, we don’t have to mathematically derive an equation for the inverse to figure out what its graph looks like. 9. Figure B-4 is a graph of the inverse of w = v 2. This is the result of using the “point reflector” scheme to modify Fig. 14-12. Again, it is not necessary to derive an equation for the inverse.

10. When the scale increments in a graph are not the same (as is the case in Fig. 14-13), we cannot use the “point reflector” scheme directly to see what the inverse graph looks like. It’s better to derive an equation for the inverse relation, and then plot its graph on the basis of that equation. We want to derive the inverse of u = t 3. We begin by transposing the values of the variables without changing their names, getting t = u3 We can take the cube root of both sides of the equation here, and we don’t run any risk of ambiguity. That’s because the original function is a bijection. Remember what that means: Every value in the domain has exactly one “mate” in the range, and vice-versa. There’s no chance for confusion or duplicity as there would be if we were dealing with an even-numbered power of u. When we take the cube root of both sides, we get ±t 1/3 = u Reversing the sense gives us u = ±t 1/3 This is the inverse of u = t 3. Figure B-5 is a graph of this relation. Note that the increments of the scales have been transposed, as well as the points in the graph, so the graph will fit neatly into the available space.
u 6 (64,4) 4 (27,3) 2 (8,2) t –40 (–27,–3) (–64,–4) –6 (–8,–2) 20 (0,0) 40 60

Figure B-5 Illustration for the solution to Prob. 10
in Chap. 14.

Chapter 15

633

Chapter 15
1. We have these two ordered pairs defining the points P and Q, respectively: P = (u1, v1) = (−1, −6) and Q = (u2, v2) = (2, 2) The slope m is equal to the difference in the dependent-variable coordinates divided by the difference in the independent-variable coordinates, or Δv/Δu. If we move along the line from P to Q, we find the slope on the basis of the ratio between the differences in the point values with the “destination” values listed first: m = (v2 − v1) / (u2 − u1) Plugging in the values v2 = 2, v1 = −6, u2 = 2, and u1 = −1, we get m = [2 − (−6)] / [2 − (−1)] = (2 + 6) / (2 + 1) = 8/3 2. We still have the same two points, defined by the same two ordered pairs. Points P and Q, respectively, are still defined by P = (u1, v1) = (−1, −6) and Q = (u2, v2) = (2, 2) If we want to go from Q to P rather than from P to Q, we must reverse the order of v1 and v2 in the numerator of the slope equation, and we must also reverse the order of u1 and u2 in the denominator. When we do that, we get m = (v1 − v2) / (u1 − u2) = (−6 − 2) / (−1 − 2) = −8 / (−3) = 8/3 The slope in either direction is equal to the difference in the v values divided by the difference in the u values, or Δv /Δu. Reversing the direction in which we move along the line simply multiplies both Δv and Δu by −1. The ratio turns out the same either way.

634 Worked-Out Solutions to Exercises: Chapters 11 to 19

3. We know the coordinates of at least one point (two, actually) and we also know the slope. Let’s use the point (2, 2) as the starting basis. We’ve determined that the slope is 8/3. The general PS equation, using u and v as the variable names rather than the familiar x and y, is v − v0 = m(u − u0) Plugging in 2 for v0, 8/3 for m, and 2 for u0, we have v − 2 = (8/3)(u − 2) That’s the PS form of the equation. 4. To get the equation in SI form, we can manipulate the PS equation we obtained in the previous solution. That equation, again, is v − 2 = (8/3)(u − 2) Using the distributive law for multiplication over subtraction on the right-hand side, we obtain v − 2 = (8/3)u − 16/3 Adding 2 to the left side, and adding 6/3 (which is equal to 2) to the right side, we get v = (8/3)u − 16/3 + 6/3 which simplifies to v = (8/3)u − 10/3 That’s the SI form of the equation. 5. The simplest possible way to graph this equation is to plot the two points we were originally given, and then draw a straight line through them. This is done in Fig. B-6. The slope, m, is 8/3 as we derived it. The v-intercept, b, is −10/3 as we derived it. 6. The first equation is in SI form. We can tell, by looking at this equation, that the slope of the line will be 1 when we graph it. The second equation can be put into SI form by considering the subtraction of s as the addition of its negative, getting t = 5 + (−s) We can apply the commutative law on the right side to get t = −s + 5 Now we can see that the slope of this line will be −1 when we graph it. In a Cartesian plane where both axes are graduated in increments of the same size, a slope of 1 corresponds to a ramp angle of 45°, and a slope of −1 corresponds to a ramp

Chapter 15 v 6 m = 8/3 4 2 Q = (2,2) u –6 –4 –2 2 4 6

635

–4 P = (–1,–6) –6

(0,–10/3) b = –10/3

Figure B-6 Illustration for the solution to Prob. 5 in
Chap. 15.

angle of −45°. That means the first line will go “uphill” at 45° as we go to the right, and the second line will go “downhill” at 45° as we go to the right. The angle between the two lines will therefore be 45 + 45, or 90°. If the increments on the s axis are not the same size as those on the t axis, then slopes of 1 and −1 will not appear as “uphill” and “downhill” 45° angles. Depending on which axis has the larger increments, both lines will be either steeper or less steep. Our advisor, who claimed that the lines would intersect at a 90° angle, will be mistaken if we draw the lines on a coordinate system having axes graduated in unequal increments. 7. To determine the point where the two lines intersect, we must find an ordered pair of the form (s, t) that satisfies both equations. Look at the SI forms of the two equations again: t=s+5 and t = −s + 5 If we add the left sides of these equations, we get t + t, which is equal to 2t. If we add the right sides, we get s + 5 + (−s) + 5, which is equal to 10. That means the sum of the two equations is 2t = 10

636 Worked-Out Solutions to Exercises: Chapters 11 to 19

Dividing through by 2, we get t = 5. That’s the t value of the intersection point. We can plug 5 in for t in either of the original equations to solve for s. Let’s use the first one. We then get 5=s+5 It’s not too difficult to tell from this equation that s = 0. Now we know that s = 0 and t = 5, so (s, t) = (0, 5) defines the point where the two lines intersect. That point lies on the t axis, because the s coordinate is equal to 0. 8. Figure B-7 shows the graphs of the two lines, based on their known slopes and t-intercepts. The intersection point is, coincidentally, at the t-intercept for both lines. The lines intersect at a 90° angle because the axes are graduated in equal increments. 9. For reference, here’s the general two-point equation we derived for a line in Cartesian coordinates: y − y1 = (x − x1)(y2 − y1) / (x2 − x1) where points are represented by ordered pairs (x1, y1) and (x2, y2). We are told that (2, 8) and (0, −4) both lie on the graph. Let’s assign x1 = 2, x2 = 0, y1 = 8, and y2 = −4. When we plug these numbers into the above equation, we get y − 8 = (x − 2)(−4 − 8) / (0 − 2)